Difference between revisions of "2005 IMO Problems/Problem 3"

Revision as of 03:12, 4 September 2025

""" Solution:

First we can easily see x 5 − x 2 x 5 + y 2 + z 2 ≥ x 4 − x x 4 + y 2 + z 2 x 5

+y

2

+z

2

x 5

−x

2

≥

x 4

+y

2

+z

2

x 4

−x

⇔ x ( x − 1 ) 2 ( x 2 + x + 1 ) ( y 2 + z 2 ) ( x 5 + y 2 + z 2 ) ( x 4 + y 2 + z 2 ) ≥ 0 ⇔ (x 5

+y

2

+z

2

)(x

4

+y

2

+z

2

x(x−1) 2

(x

2

+x+1)(y

2

+z

2

≥0 which is true for

x , y , z x,y,z and z z are positive real numbers.

It's enough to prove: x 4 − x x 4 + y 2 + z 2 + y 4 − y y 4 + z 2 + x 2 + z 4 − z z 4 + x 2 + y 2 ≥ 0 x 4

+y

2

+z

2

x 4

−x

y 4

+z

2

+x

2

y 4

−y

z 4

+x

2

+y

2

z 4

−z

≥0

Or x 4 x 4 + y 2 + z 2 + y 4 y 4 + z 2 + x 2 + z 4 z 4 + x 2 + y 2 ≥ x x 4 + y 2 + z 2 + y y 4 + z 2 + x 2 + z z 4 + x 2 + y 2 x 4

+y

2

+z

2

x 4

y 4

+z

2

+x

2

y 4

z 4

+x

2

+y

2

z 4

≥

x 4

+y

2

+z

2

x

y 4

+z

2

+x

2

y

z 4

+x

2

+y

2

z

Let t = x + y + z → t ≥ 3 ( x y z ) 1 / 3 ≥ 3 t=x+y+z→t≥3(xyz) 1/3

≥3 and

x y + y z + z x ≤ t 2 3 ≤ x 2 + y 2 + z 2 xy+yz+zx≤ 3 t 2

≤x

2

+y

2

+z

2

By C-S: ( x 2 + y 2 + z 2 ) 2 ≤ ( x 4 + y 2 + z 2 ) ( 1 + y 2 + z 2 ) → x x 4 + y 2 + z 2 ≤ x ( 1 + y 2 + z 2 ) ( x 2 + y 2 + z 2 ) 2 (x 2

+y

2

+z

2

≤(x

4

+y

2

+z

2

)(1+y

2

+z

2

)→

x 4

+y

2

+z

2

x

≤

(x 2

+y

2

+z

2

x(1+y 2

+z

2

⇒ ⇒ RHS ≤ x ( 1 + y 2 + z 2 ) + y ( 1 + z 2 + x 2 ) + z ( 1 + x 2 + y 2 ) ( x 2 + y 2 + z 2 ) 2 = x + y + z + ( x y + y z + z x ) ( x + y + z ) − 3 x y z ( x 2 + y 2 + z 2 ) 2 ≤ (x 2

+y

2

+z

2

x(1+y 2

+z

2

)+y(1+z

2

+x

2

)+z(1+x

2

+y

2

(x 2

+y

2

+z

2

x+y+z+(xy+yz+zx)(x+y+z)−3xyz

≤ t + t 3 3 t 4 = 3 t 3 + 9 t − 27 t 4 ≤ 1 ⇔ ( t − 3 ) ( t 3 − 9 ) ≥ 0 ≤ t 4

t+ 3 t 3

t 4

3t 3

+9t−27

≤1⇔(t−3)(t

3

−9)≥0

which is true for positive real numbers $x, y, z$ .

It's enough to prove:

x 4 − x x 4 + y 2 + z 2 + y 4 − y y 4 + z 2 + x 2 + z 4 − z z 4 + x 2 + y 2 ≥ 0. x 4

+y

2

+z

2

x 4

−x

y 4

+z

2

+x

2

y 4

−y

z 4

+x

2

+y

2

z 4

−z

≥0.

Or equivalently,

x 4 x 4 + y 2 + z 2 + y 4 y 4 + z 2 + x 2 + z 4 z 4 + x 2 + y 2 ≥ x x 4 + y 2 + z 2 + y y 4 + z 2 + x 2 + z z 4 + x 2 + y 2 . x 4

+y

2

+z

2

x 4

y 4

+z

2

+x

2

y 4

z 4

+x

2

+y

2

z 4

≥

x 4

+y

2

+z

2

x

y 4

+z

2

+x

2

y

z 4

+x

2

+y

2

z

Let $t = x + y + z$ . Then by AM–GM, $t \geq 3(xyz)^{1/3} \geq 3$ and $xy + yz + zx \leq \frac{t^2}{3} \leq x^2 + y^2 + z^2$ .

By Cauchy–Schwarz:

( x 2 + y 2 + z 2 ) 2 ≤ ( x 4 + y 2 + z 2 ) ( 1 + y 2 + z 2 ) ⇒ x x 4 + y 2 + z 2 ≤ x ( 1 + y 2 + z 2 ) ( x 2 + y 2 + z 2 ) 2 . (x 2

+y

2

+z

2

≤(x

4

+y

2

+z

2

)(1+y

2

+z

2

)⇒

x 4

+y

2

+z

2

x

≤

(x 2

+y

2

+z

2

x(1+y 2

+z

2

Hence, the right-hand side (RHS) satisfies \begin{align*} \text{RHS} &\leq \frac{x(1 + y^2 + z^2) + y(1 + z^2 + x^2) + z(1 + x^2 + y^2)}{(x^2 + y^2 + z^2)^2} \ &= \frac{x+y+z + (xy+yz+zx)(x+y+z) - 3xyz}{(x^2 + y^2 + z^2)^2}. \end{align*} Since $x^2+y^2+z^2 \geq \frac{t^2}{3}$ and $xy+yz+zx \leq \frac{t^2}{3}$ , we have

RHS ≤ t + t 3 3 t 4 / 9 = 9 t 4 ( t + t 3 3 ) = 9 t 4 ⋅ 3 t + t 3 3 = 3 ( t 3 + 3 t ) t 4 = 3 t 3 + 9 t t 4 . RHS≤ t 4

/9

t+ 3 t 3

t 4

9

(t+

3 t 3

)=

t 4

9

⋅

3 3t+t 3

t 4

3(t 3

+3t)

t 4

3t 3

+9t

Wait, the original says:

≤ t + t 3 3 t 4 = 3 t 3 + 9 t − 27 t 4 ≤ 1 ⇔ ( t − 3 ) ( t 3 − 9 ) ≥ 0 ≤ t 4

t+ 3 t 3

t 4

3t 3

+9t−27

≤1⇔(t−3)(t

3

−9)≥0

But there is a discrepancy: The step from the previous expression to the next is not clear. Actually, the original says:

≤ t + t 3 3 t 4 = 3 t 3 + 9 t − 27 t 4 ≤ t 4

t+ 3 t 3

t 4

3t 3

+9t−27

That seems to have an error: Actually, the original might have intended: We have:

x + y + z + ( x y + y z + z x ) ( x + y + z ) − 3 x y z ( x 2 + y 2 + z 2 ) 2 ≤ t + ( t 2 / 3 ) t t 4 / 9 = t + t 3 / 3 t 4 / 9 = 9 t + t 3 / 3 t 4 = 9 t + 3 t 3 t 4 = 3 t 3 + 9 t t 4 . (x 2

+y

2

+z

2

x+y+z+(xy+yz+zx)(x+y+z)−3xyz

≤

t 4

/9

t+(t 2

/3)t

t 4

/9

t+t 3

/3

=9

t 4

t+t 3

/3

t 4

9t+3t 3

t 4

3t 3

+9t

But then the original writes:

≤ t + t 3 3 t 4 = 3 t 3 + 9 t − 27 t 4 ≤ t 4

t+ 3 t 3

t 4

3t 3

+9t−27

That is inconsistent. Possibly there is a misprint. Actually, the original then says:

3 t 3 + 9 t − 27 t 4 ≤ 1 ⇔ ( t − 3 ) ( t 3 − 9 ) ≥ 0. t 4

3t 3

+9t−27

≤1⇔(t−3)(t

3

−9)≥0.

But if we had $\frac{3t^3+9t}{t^4}$ , then we would need to show that $\frac{3t^3+9t}{t^4} \leq 1$ for $t \geq 3$ , which is $3t^3+9t \leq t^4$ or $t^4 - 3t^3 - 9t \geq 0$ , i.e., $t(t^3-3t^2-9) \geq 0$ . That is not the same as $(t-3)(t^3-9) \geq 0$ . So there is a discrepancy.

Let's re-read the original: "RHS $\leq \frac{x+y+z+(xy+yz+zx)(x+y+z)-3xyz}{(x^2 + y^2 + z^2)^2} \leq \frac{t + \frac{t^3}{3}}{t^4} = \frac{3t^3 + 9t - 27}{t^4} \leq 1$ "

Wait, the original has:

≤ t + t 3 3 t 4 = 3 t 3 + 9 t − 27 t 4 ≤ t 4

t+ 3 t 3

t 4

3t 3

+9t−27

But $\frac{t + \frac{t^3}{3}}{t^4} = \frac{3t + t^3}{3t^4} = \frac{t^3+3t}{3t^4}$ , not $\frac{3t^3+9t-27}{t^4}$ . So there is a mistake.

Maybe it is:

≤ t + t 3 3 − 3 ( x 2 + y 2 + z 2 ) 2 ≤ t + t 3 3 − 3 t 4 / 9 = 9 ( t + t 3 / 3 − 3 ) t 4 = 9 t + 3 t 3 − 27 t 4 = 3 t 3 + 9 t − 27 t 4 ≤ (x 2

+y

2

+z

2

t+ 3 t 3

−3

≤

t 4

/9

t+ 3 t 3

−3

t 4

9(t+t 3

/3−3)

t 4

9t+3t 3

−27

t 4

3t 3

+9t−27

because we also subtract 3xyz? Actually, note that the numerator is: x+y+z + (xy+yz+zx)(x+y+z) - 3xyz. And we have xyz >= 1, so -3xyz <= -3. So we can bound: x+y+z + (xy+yz+zx)(x+y+z) - 3xyz <= t + (t^2/3)*t - 3 = t + t^3/3 - 3. And then denominator: (x^2+y^2+z^2)^2 >= (t^2/3)^2 = t^4/9. So indeed: RHS <= (t + t^3/3 - 3) / (t^4/9) = 9(t + t^3/3 - 3)/t^4 = (9t + 3t^3 - 27)/t^4 = (3t^3+9t-27)/t^4. So that step is correct:

RHS ≤ 3 t 3 + 9 t − 27 t 4 . RHS≤ t 4

3t 3

+9t−27

Then we want to show that this is ≤ 1, i.e.,

3 t 3 + 9 t − 27 t 4 ≤ 1 ⇔ 3 t 3 + 9 t − 27 ≤ t 4 ⇔ t 4 − 3 t 3 − 9 t + 27 ≥ 0. t 4

3t 3

+9t−27

≤1⇔3t

3

+9t−27≤t

4

⇔t

4

−3t

3

−9t+27≥0.

Factor: t^4 - 3t^3 - 9t + 27 = (t-3)(t^3 - 9) ? Let's check: (t-3)(t^3 - 9) = t^4 - 9t - 3t^3 + 27 = t^4 - 3t^3 - 9t + 27. Yes. So we need (t-3)(t^3-9) >= 0. For t>=3, t-3>=0 and t^3-9>=0, so indeed it holds. So RHS ≤ 1.

Then the original says: "By AM-GM: LHS ≥ ...". But careful: The LHS here is the left-hand side of the inequality we want to prove? Actually, we want to prove that LHS (which is the sum of fractions with x^4) is at least 1. So we need to show LHS ≥ 1.

The original then says: LHS ≥ (x^3+y^3+z^3)^2 / (x^6+y^6+z^6+2(x^2y^2+y^2z^2+z^2x^2)) ≥ 1. That is by Cauchy-Schwarz? Actually, by Titu's lemma (Cauchy-Schwarz in Engel form): ∑ x^4/(x^4+y^2+z^2) ≥ (∑ x^2)^2 / ∑ (x^4+y^2+z^2) = (x^2+y^2+z^2)^2 / (∑ x^4 + 2∑ x^2) but careful: Actually, the denominators are different: The first term denominator is x^4+y^2+z^2, so if we sum over cyclic, then the denominator sum becomes (x^4+y^4+z^4) + (y^2+z^2) + (z^2+x^2) + (x^2+y^2) = x^4+y^4+z^4 + 2(x^2+y^2+z^2). So then by Cauchy, we have: LHS = ∑ x^4/(x^4+y^2+z^2) ≥ (x^2+y^2+z^2)^2 / (x^4+y^4+z^4+2(x^2+y^2+z^2)). But the original writes: LHS ≥ (x^3+y^3+z^3)^2 / (x^6+y^6+z^6+2(x^2y^2+y^2z^2+z^2x^2)). That is different: They have x^3 in numerator squared, and denominator: x^6+y^6+z^6+2(x^2y^2+...). Actually, note that (x^3)^2 = x^6. So it is similar but with squares? Actually, if we set a = x^2, then a^2 = x^4. So it's not exactly.

Wait, maybe they used: By Cauchy, (∑ x^4/(x^4+y^2+z^2)) (∑ x^4(x^4+y^2+z^2)) ≥ (∑ x^4)^2. But then ∑ x^4(x^4+y^2+z^2) = ∑ (x^8 + x^4y^2 + x^4z^2) = ∑ x^8 + ∑ x^4y^2 + ∑ x^4z^2. That is not the same as x^6+y^6+z^6+2(x^2y^2+...)? Actually, x^6? No.

Alternatively, they might have used the inequality: ∑ x^4/(x^4+y^2+z^2) ≥ (∑ x^2)^2 / (∑ x^4 + 2∑ x^2) but then they claim that is ≥ 1 if (x^2+y^2+z^2)^2 ≥ x^4+y^4+z^4+2(x^2+y^2+z^2) which is not necessarily true.

Maybe they used a different approach: They said: LHS ≥ (x^3+y^3+z^3)^2 / (x^6+y^6+z^6+2(x^2y^2+y^2z^2+z^2x^2)). That would come from Cauchy if we write: (∑ x^4/(x^4+y^2+z^2)) (∑ (x^4+y^2+z^2)) ≥ (∑ x^2)^2? Not exactly.

Wait, consider: We want to lower bound LHS = ∑ x^4/(x^4+y^2+z^2). By Cauchy, (∑ x^4/(x^4+y^2+z^2)) (∑ x^4(x^4+y^2+z^2)) ≥ (∑ x^4)^2. But then that gives LHS ≥ (∑ x^4)^2 / (∑ (x^8 + x^4y^2+x^4z^2)) = (∑ x^4)^2 / (∑ x^8 + ∑ x^4y^2+∑ x^4z^2). And note that ∑ x^8 = x^8+y^8+z^8, and ∑ x^4y^2+∑ x^4z^2 = ∑ x^4(y^2+z^2) = ∑ x^4((x^2+y^2+z^2)-x^2) = (x^2+y^2+z^2)∑ x^4 - ∑ x^6. So then denominator = ∑ x^8 + (x^2+y^2+z^2)∑ x^4 - ∑ x^6. That is not the same as x^6+y^6+z^6+2(x^2y^2+y^2z^2+z^2x^2) unless further conditions.

Alternatively, maybe they used: By Titu's lemma: ∑ x^4/(x^4+y^2+z^2) ≥ (∑ x^2)^2 / (∑ (x^4+y^2+z^2)) = (x^2+y^2+z^2)^2 / (∑ x^4 + 2∑ x^2). But then they claim that is ≥ 1 if (x^2+y^2+z^2)^2 ≥ ∑ x^4+2∑ x^2, i.e., 2(x^2y^2+y^2z^2+z^2x^2) ≥ 2(x^2+y^2+z^2) or x^2y^2+y^2z^2+z^2x^2 ≥ x^2+y^2+z^2, which is not necessarily true.

Given that the original text says: "LHS ≥ \frac{(x^3 + y^3 + z^3)^2}{x^6 + y^6 + z^6 + 2(x^2y^2 + y^2z^2 + z^2x^2)} \geq 1" Maybe it is actually: LHS = ∑ x^4/(x^4+y^2+z^2) and they use the substitution: Let a = x^2, b = y^2, c = z^2. Then LHS = ∑ a^2/(a^2+b+c). And then by Cauchy, (∑ a^2/(a^2+b+c)) (∑ a^2(a^2+b+c)) ≥ (∑ a^2)^2. But ∑ a^2(a^2+b+c) = ∑ (a^4 + a^2b + a^2c) = ∑ a^4 + ∑ a^2(b+c) = ∑ a^4 + ∑ a^2((a+b+c)-a) = ∑ a^4 + (a+b+c)∑ a^2 - ∑ a^3. So then LHS ≥ (∑ a^2)^2 / (∑ a^4 + (a+b+c)∑ a^2 - ∑ a^3). That doesn't simplify to the given.

Alternatively, they might have used the inequality: a^2/(a^2+b+c) ≥ (a^3)^2/(a^6 + a^3(b+c))? That is not standard.

Wait, the given expression: \frac{(x^3+y^3+z^3)^2}{x^6+y^6+z^6+2(x^2y^2+y^2z^2+z^2x^2)} Notice that x^6+y^6+z^6+2(x^2y^2+y^2z^2+z^2x^2) = (x^2+y^2+z^2)^3? Actually, (x^2+y^2+z^2)^3 = x^6+y^6+z^6+3(x^2y^2(x^2+y^2)+... so no.

Maybe it's a misprint: They might have meant: LHS ≥ \frac{(x^2+y^2+z^2)^2}{x^4+y^4+z^4+2(x^2+y^2+z^2)}. But then they claim that is ≥ 1 if (x^2+y^2+z^2)^2 ≥ x^4+y^4+z^4+2(x^2+y^2+z^2) which is equivalent to 2(x^2y^2+y^2z^2+z^2x^2) ≥ 2(x^2+y^2+z^2) or x^2y^2+y^2z^2+z^2x^2 ≥ x^2+y^2+z^2. That is not generally true.

Given the subsequent steps: They then say: ≥ 1 ⇔ x^3y^3+y^3z^3+z^3x^3 ≥ x^2y^2+y^2z^2+z^2x^2 ⇔ a^3+b^3+c^3 ≥ a^2+b^2+c^2, where a=xy, b=yz, c=zx. So indeed, they are comparing: (x^3+y^3+z^3)^2 and x^6+y^6+z^6+2(x^2y^2+y^2z^2+z^2x^2). Note that (x^3+y^3+z^3)^2 = x^6+y^6+z^6+2(x^3y^3+x^3z^3+y^3z^3). So the inequality \frac{(x^3+y^3+z^3)^2}{x^6+y^6+z^6+2(x^2y^2+y^2z^2+z^2x^2)} \geq 1 is equivalent to x^6+y^6+z^6+2(x^3y^3+x^3z^3+y^3z^3) \geq x^6+y^6+z^6+2(x^2y^2+y^2z^2+z^2x^2) i.e., x^3y^3+x^3z^3+y^3z^3 \geq x^2y^2+y^2z^2+z^2x^2. So that step is correct: They claim that LHS (the sum of fractions) is at least that fraction, and then that fraction is at least 1 if x^3y^3+... >= x^2y^2+... So then they set a=xy, b=yz, c=zx, so that a^3+b^3+c^3 >= a^2+b^2+c^2. And then they note that abc = x^2y^2z^2, and since x,y,z are positive with xyz>=1, then abc>=1. And then they use Holder: (a^2+b^2+c^2)^3 <= 3(a^3+b^3+c^3)^2 <= (a^3+b^3+c^3)^3? Actually, they write: By Holder: (a^2+b^2+c^2)^3 <= 3(a^3+b^3+c^3)^2 <= (a^3+b^3+c^3)^3 ⇒ a^2+b^2+c^2 <= a^3+b^3+c^3. Wait, check: Holder says: (a^3+b^3+c^3)(1+1+1)(1+1+1) >= (a+b+c)^3, but that's not directly. They claim: (a^2+b^2+c^2)^3 <= 3(a^3+b^3+c^3)^2. Is that true? By power mean, ( (a^3+b^3+c^3)/3 )^(1/3) >= ( (a^2+b^2+c^2)/3 )^(1/2) so (a^3+b^3+c^3) >= 3^{-1/2} (a^2+b^2+c^2)^(3/2)? That gives (a^3+b^3+c^3)^2 >= (1/3)(a^2+b^2+c^2)^3, so actually (a^2+b^2+c^2)^3 <= 3(a^3+b^3+c^3)^2. So that is true. Then they claim: 3(a^3+b^3+c^3)^2 <= (a^3+b^3+c^3)^3, which would require a^3+b^3+c^3 >= 3. And since a,b,c are such that abc>=1, by AM-GM, a^3+b^3+c^3 >= 3(abc) >= 3. So indeed, a^3+b^3+c^3 >= 3. Then we have: (a^2+b^2+c^2)^3 <= 3(a^3+b^3+c^3)^2 <= (a^3+b^3+c^3)^3, so taking cube roots: a^2+b^2+c^2 <= a^3+b^3+c^3. So that step is valid.

But then the claim that LHS >= that fraction is not justified in the text. They simply say "By AM-GM:" and then write that inequality. Possibly they meant by Cauchy-Schwarz? Let's check: We want to show: ∑ x^4/(x^4+y^2+z^2) >= (x^3+y^3+z^3)^2 / (x^6+y^6+z^6+2(x^2y^2+y^2z^2+z^2x^2)). This is equivalent to: (∑ x^4/(x^4+y^2+z^2)) (x^6+y^6+z^6+2(x^2y^2+y^2z^2+z^2x^2)) >= (x^3+y^3+z^3)^2. Now, by Cauchy, we have: (∑ x^4/(x^4+y^2+z^2)) (∑ x^4(x^4+y^2+z^2)) >= (∑ x^4)^2. But we need (∑ x^3)^2 on the right. So if we can show that (∑ x^4)^2 >= (∑ x^3)^2, i.e., ∑ x^4 >= ∑ x^3, that is not always true. Alternatively, if we use the weights differently: We want to relate x^4(x^4+y^2+z^2) to something like x^6? Not sure.

Maybe they used the following: x^4/(x^4+y^2+z^2) >= x^6/(x^6+x^2y^2+x^2z^2) because multiply numerator and denominator by x^2 gives: x^6/(x^6+x^2y^2+x^2z^2). And then sum cyclically: LHS >= ∑ x^6/(x^6+x^2y^2+x^2z^2). Then by Cauchy, ∑ x^6/(x^6+x^2y^2+x^2z^2) >= (∑ x^3)^2 / (∑ x^6+∑ x^2y^2+∑ x^2z^2) = (∑ x^3)^2 / (∑ x^6+2∑ x^2y^2). That is exactly the claimed lower bound. And the inequality x^4/(x^4+y^2+z^2) >= x^6/(x^6+x^2y^2+x^2z^2) is equivalent to cross multiply: x^4(x^6+x^2y^2+x^2z^2) >= x^6(x^4+y^2+z^2) => x^10+x^6y^2+x^6z^2 >= x^10+x^6y^2+x^6z^2, which is equality! Actually, check: x^4/(x^4+y^2+z^2) >= x^6/(x^6+x^2y^2+x^2z^2) Multiply both sides by positive denominators: x^4(x^6+x^2y^2+x^2z^2) >= x^6(x^4+y^2+z^2) => x^10 + x^6y^2 + x^6z^2 >= x^10 + x^6y^2 + x^6z^2. So it's actually equality. So indeed, x^4/(x^4+y^2+z^2) = x^6/(x^6+x^2y^2+x^2z^2). Because multiply numerator and denominator by x^2 gives that provided x>0. So we have: LHS = ∑ x^4/(x^4+y^2+z^2) = ∑ x^6/(x^6+x^2y^2+x^2z^2). Then by Cauchy-Schwarz, ∑ x^6/(x^6+x^2y^2+x^2z^2) ≥ (∑ x^3)^2 / (∑ x^6 + ∑ x^2y^2+∑ x^2z^2) = (∑ x^3)^2 / (∑ x^6+2∑ x^2y^2). So that step is justified.

So we will include that step.

Now, let's write the LaTeX code accordingly.

I'll structure it as:

\documentclass{article} \usepackage{amsmath, amssymb} \begin{document}

\section*{Solution}

First, we can easily see that

x 5 − x 2 x 5 + y 2 + z 2 ≥ x 4 − x x 4 + y 2 + z 2 x 5

+y

2

+z

2

x 5

−x

2

≥

x 4

+y

2

+z

2

x 4

−x

since

x 5 − x 2 x 5 + y 2 + z 2 − x 4 − x x 4 + y 2 + z 2 = x ( x − 1 ) 2 ( x 2 + x + 1 ) ( y 2 + z 2 ) ( x 5 + y 2 + z 2 ) ( x 4 + y 2 + z 2 ) ≥ 0 , x 5

+y

2

+z

2

x 5

−x

2

−

x 4

+y

2

+z

2

x 4

−x

(x 5

+y

2

+z

2

)(x

4

+y

2

+z

2

x(x−1) 2

(x

2

+x+1)(y

2

+z

2

≥0,

which holds for all positive real numbers $x, y, z$ .

Thus, it is enough to prove that

x 4 − x x 4 + y 2 + z 2 + y 4 − y y 4 + z 2 + x 2 + z 4 − z z 4 + x 2 + y 2 ≥ 0. x 4

+y

2

+z

2

x 4

−x

y 4

+z

2

+x

2

y 4

−y

z 4

+x

2

+y

2

z 4

−z

≥0.

Rewriting, we need to show

x 4 x 4 + y 2 + z 2 + y 4 y 4 + z 2 + x 2 + z 4 z 4 + x 2 + y 2 ≥ x x 4 + y 2 + z 2 + y y 4 + z 2 + x 2 + z z 4 + x 2 + y 2 . x 4

+y

2

+z

2

x 4

y 4

+z

2

+x

2

y 4

z 4

+x

2

+y

2

z 4

≥

x 4

+y

2

+z

2

x

y 4

+z

2

+x

2

y

z 4

+x

2

+y

2

z

Let $t = x + y + z$ . By the AM–GM inequality, we have $t \geq 3\sqrt[3]{xyz} \geq 3$ , and also $xy + yz + zx \leq \frac{t^2}{3} \leq x^2 + y^2 + z^2$ .

Now, estimate the right-hand side (RHS). By the Cauchy–Schwarz inequality,

( x 2 + y 2 + z 2 ) 2 ≤ ( x 4 + y 2 + z 2 ) ( 1 + y 2 + z 2 ) , (x 2

+y

2

+z

2

≤(x

4

+y

2

+z

2

)(1+y

2

+z

2

),

so

x x 4 + y 2 + z 2 ≤ x ( 1 + y 2 + z 2 ) ( x 2 + y 2 + z 2 ) 2 . x 4

+y

2

+z

2

x

≤

(x 2

+y

2

+z

2

x(1+y 2

+z

2

Cyclically summing, we get \begin{align*} \text{RHS} &\leq \frac{x(1+y^2+z^2) + y(1+z^2+x^2) + z(1+x^2+y^2)}{(x^2+y^2+z^2)^2} \ &= \frac{x+y+z + (xy+yz+zx)(x+y+z) - 3xyz}{(x^2+y^2+z^2)^2}. \end{align*} Since $xyz \geq 1$ , we have $-3xyz \leq -3$ . Also, $xy+yz+zx \leq \frac{t^2}{3}$ and $x^2+y^2+z^2 \geq \frac{t^2}{3}$ . Hence, \begin{align*} \text{RHS} &\leq \frac{t + \frac{t^2}{3} \cdot t - 3}{(t^2/3)^2} = \frac{t + \frac{t^3}{3} - 3}{t^4/9} = \frac{9(t + t^3/3 - 3)}{t^4} = \frac{9t + 3t^3 - 27}{t^4} = \frac{3t^3 + 9t - 27}{t^4}. \end{align*} We claim that

3 t 3 + 9 t − 27 t 4 ≤ 1. t 4

3t 3

+9t−27

≤1.

This is equivalent to

3 t 3 + 9 t − 27 ≤ t 4 ⇔ t 4 − 3 t 3 − 9 t + 27 ≥ 0. 3t 3

+9t−27≤t

4

⇔t

4

−3t

3

−9t+27≥0.

Factoring, we have

t 4 − 3 t 3 − 9 t + 27 = ( t − 3 ) ( t 3 − 9 ) ≥ 0 , t 4

−3t

3

−9t+27=(t−3)(t

3

−9)≥0,

which holds for $t \geq 3$ . Therefore, RHS $\leq 1$ .

Now, estimate the left-hand side (LHS). Note that for $x > 0$ , we have

x 4 x 4 + y 2 + z 2 = x 6 x 6 + x 2 y 2 + x 2 z 2 . x 4

+y

2

+z

2

x 4

x 6

+x

2

+x

2

x 6

Thus,

LHS = ∑ x 4 x 4 + y 2 + z 2 = ∑ x 6 x 6 + x 2 y 2 + x 2 z 2 . LHS=∑ x 4

+y

2

+z

2

x 4

=∑

x 6

+x

2

+x

2

x 6

By the Cauchy–Schwarz inequality,

( ∑ x 6 x 6 + x 2 y 2 + x 2 z 2 ) ( ∑ ( x 6 + x 2 y 2 + x 2 z 2 ) ) ≥ ( ∑ x 3 ) 2 . (∑ x 6

+x

2

+x

2

x 6

)(∑(x

6

+x

2

+x

2

))≥(∑x

3

2

But

∑ ( x 6 + x 2 y 2 + x 2 z 2 ) = ∑ x 6 + ∑ x 2 y 2 + ∑ x 2 z 2 = ∑ x 6 + 2 ∑ x 2 y 2 . ∑(x 6

+x

2

+x

2

)=∑x

6

+∑x

2

+∑x

2

=∑x

6

+2∑x

2

Hence,

LHS ≥ ( x 3 + y 3 + z 3 ) 2 x 6 + y 6 + z 6 + 2 ( x 2 y 2 + y 2 z 2 + z 2 x 2 ) . LHS≥ x 6

+y

6

+z

6

+2(x

2

+y

2

+z

2

(x 3

+y

3

+z

3

2

We now show that this expression is at least 1. That is, we need

( x 3 + y 3 + z 3 ) 2 ≥ x 6 + y 6 + z 6 + 2 ( x 2 y 2 + y 2 z 2 + z 2 x 2 ) . (x 3

+y

3

+z

3

2

≥x

6

+y

6

+z

6

+2(x

2

+y

2

+z

2

).

Expanding the left-hand side:

( x 3 + y 3 + z 3 ) 2 = x 6 + y 6 + z 6 + 2 ( x 3 y 3 + x 3 z 3 + y 3 z 3 ) . (x 3

+y

3

+z

3

2

=x

6

+y

6

+z

6

+2(x

3

+x

3

+y

3

).

So the inequality becomes

x 3 y 3 + x 3 z 3 + y 3 z 3 ≥ x 2 y 2 + y 2 z 2 + z 2 x 2 . x 3

3

+x

3

+y

3

≥x

2

+y

2

+z

2

Let $a = xy$ , $b = yz$ , $c = zx$ . Then the inequality is equivalent to

a 3 + b 3 + c 3 ≥ a 2 + b 2 + c 2 . a 3

+b

3

+c

3

≥a

2

+b

2

+c

2

Note that $abc = x^2y^2z^2 \geq 1$ since $xyz \geq 1$ . Also, by the AM–GM inequality, $a^3 + b^3 + c^3 \geq 3\sqrt[3]{a^3b^3c^3} = 3abc \geq 3$ .

Now, by the power mean inequality, we have

( a 3 + b 3 + c 3 3 ) 1 / 3 ≥ ( a 2 + b 2 + c 2 3 ) 1 / 2 , ( 3 a 3

+b

3

+c

3

1/3

≥(

3 a 2

+b

2

+c

2

1/2

which implies

( a 3 + b 3 + c 3 ) 2 ≥ 1 3 ( a 2 + b 2 + c 2 ) 3 . (a 3

+b

3

+c

3

2

≥

3 1

(a

2

+b

2

+c

2

3

That is,

( a 2 + b 2 + c 2 ) 3 ≤ 3 ( a 3 + b 3 + c 3 ) 2 . (a 2

+b

2

+c

2

3

≤3(a

3

+b

3

+c

3

2

Since $a^3 + b^3 + c^3 \geq 3$ , we have

3 ( a 3 + b 3 + c 3 ) 2 ≤ ( a 3 + b 3 + c 3 ) 3 . 3(a 3

+b

3

+c

3

2

≤(a

3

+b

3

+c

3

Combining, we get

( a 2 + b 2 + c 2 ) 3 ≤ ( a 3 + b 3 + c 3 ) 3 , (a 2

+b

2

+c

2

3

≤(a

3

+b

3

+c

3

so taking cube roots yields

a 2 + b 2 + c 2 ≤ a 3 + b 3 + c 3 , a 2

+b

2

+c

2

≤a

3

+b

3

+c

3

as desired. Therefore, LHS $\geq 1$ .

Since we have shown that LHS $\geq 1$ and RHS $\leq 1$ , it follows that

x 4 x 4 + y 2 + z 2 + y 4 y 4 + z 2 + x 2 + z 4 z 4 + x 2 + y 2 ≥ x x 4 + y 2 + z 2 + y y 4 + z 2 + x 2 + z z 4 + x 2 + y 2 , x 4

+y

2

+z

2

x 4

y 4

+z

2

+x

2

y 4

z 4

+x

2

+y

2

z 4

≥

x 4

+y

2

+z

2

x

y 4

+z

2

+x

2

y

z 4

+x

2

+y

2

z

and hence the original inequality holds.

Equality occurs when $x = y = z = 1$ .

\end{document}

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Difference between revisions of "2005 IMO Problems/Problem 3"

Revision as of 03:12, 4 September 2025

@@ Line 1: / Line 1: @@
-==Problem==
+"""
+Solution:
-Let <math>x, y, z > 0</math> satisfy <math>xyz\ge 1</math>. Prove that <cmath>\frac{x^5-x^2}{x^5+y^2+z^2} + \frac{y^5-y^2}{x^2+y^5+z^2} + \frac{z^5-z^2}{x^2+y^2+z^5} \ge 0.</cmath>
+First we can easily see
+x
+−
+x
+x
++
+y
++
+z
+≥
+x
+−
+x
+x
++
+y
++
+z
+x
+ +y
+ +z
+x
+ −x
+ ≥
+x
+ +y
+ +z
+x
+ −x
+⇔
+x
+(
+x
+−
+)
+(
+x
++
+x
++
+)
+(
+y
++
+z
+)
+(
+x
++
+y
++
+z
+)
+(
+x
++
+y
++
+z
+)
+≥
+⇔
+(x
+ +y
+ +z
+ )(x
+ +y
+ +z
+ )
+x(x−1)
+ (x
+ +x+1)(y
+ +z
+ )
+ ≥0 which is true for
+x
+,
+y
+,
+z
+x,y,z and
+z
+z are positive real numbers.
-==Solution==
+It's enough to prove:
-{{solution}}
+x
+−
+x
+x
++
+y
++
+z
++
+y
+−
+y
+y
++
+z
++
+x
++
+z
+−
+z
+z
++
+x
++
+y
+≥
+x
+ +y
+ +z
+x
+ −x
+ +
+y
+ +z
+ +x
+y
+ −y
+ +
+z
+ +x
+ +y
+z
+ −z
+ ≥0
-==See Also==
+Or
+x
+x
++
+y
++
+z
++
+y
+y
++
+z
++
+x
++
+z
+z
++
+x
++
+y
+≥
+x
+x
++
+y
++
+z
++
+y
+y
++
+z
++
+x
++
+z
+z
++
+x
++
+y
+x
+ +y
+ +z
+x
+ +
+y
+ +z
+ +x
+y
+ +
+z
+ +x
+ +y
+z
+ ≥
+x
+ +y
+ +z
+x
+ +
+y
+ +z
+ +x
+y
+ +
+z
+ +x
+ +y
+z
-{{IMO box|year=2005|num-b=2|num-a=4}}
+Let
+t
+=
+x
++
+y
++
+z
+→
+t
+≥
+(
+x
+y
+z
+)
+/
+≥
+t=x+y+z→t≥3(xyz)
+/3
+ ≥3 and
+x
+y
++
+y
+z
++
+z
+x
+≤
+t
+≤
+x
++
+y
++
+z
+xy+yz+zx≤
+t
+ ≤x
+ +y
+ +z
+By C-S:
+(
+x
++
+y
++
+z
+)
+≤
+(
+x
++
+y
++
+z
+)
+(
++
+y
++
+z
+)
+→
+x
+x
++
+y
++
+z
+≤
+x
+(
++
+y
++
+z
+)
+(
+x
++
+y
++
+z
+)
+(x
+ +y
+ +z
+ )
+ ≤(x
+ +y
+ +z
+ )(1+y
+ +z
+ )→
+x
+ +y
+ +z
+x
+ ≤
+(x
+ +y
+ +z
+ )
+x(1+y
+ +z
+ )
+⇒
+⇒ RHS
+≤
+x
+(
++
+y
++
+z
+)
++
+y
+(
++
+z
++
+x
+)
++
+z
+(
++
+x
++
+y
+)
+(
+x
++
+y
++
+z
+)
+=
+x
++
+y
++
+z
++
+(
+x
+y
++
+y
+z
++
+z
+x
+)
+(
+x
++
+y
++
+z
+)
+−
+x
+y
+z
+(
+x
++
+y
++
+z
+)
+≤
+(x
+ +y
+ +z
+ )
+x(1+y
+ +z
+ )+y(1+z
+ +x
+ )+z(1+x
+ +y
+ )
+ =
+(x
+ +y
+ +z
+ )
+x+y+z+(xy+yz+zx)(x+y+z)−3xyz
+≤
+t
++
+t
+t
+=
+t
++
+t
+−
+t
+≤
+⇔
+(
+t
+−
+)
+(
+t
+−
+)
+≥
+≤
+t
+t+
+t
+ =
+t
+t
+ +9t−27
+ ≤1⇔(t−3)(t
+ −9)≥0
+which is true for positive real numbers <math>x, y, z</math>.
+It's enough to prove:
+x
+−
+x
+x
++
+y
++
+z
++
+y
+−
+y
+y
++
+z
++
+x
++
+z
+−
+z
+z
++
+x
++
+y
+≥
+.
+x
+ +y
+ +z
+x
+ −x
+ +
+y
+ +z
+ +x
+y
+ −y
+ +
+z
+ +x
+ +y
+z
+ −z
+ ≥0.
+Or equivalently,
+x
+x
++
+y
++
+z
++
+y
+y
++
+z
++
+x
++
+z
+z
++
+x
++
+y
+≥
+x
+x
++
+y
++
+z
++
+y
+y
++
+z
++
+x
++
+z
+z
++
+x
++
+y
+.
+x
+ +y
+ +z
+x
+ +
+y
+ +z
+ +x
+y
+ +
+z
+ +x
+ +y
+z
+ ≥
+x
+ +y
+ +z
+x
+ +
+y
+ +z
+ +x
+y
+ +
+z
+ +x
+ +y
+z
+ .
+Let <math>t = x + y + z</math>. Then by AM–GM, <math>t \geq 3(xyz)^{1/3} \geq 3</math> and <math>xy + yz + zx \leq \frac{t^2}{3} \leq x^2 + y^2 + z^2</math>.
+By Cauchy–Schwarz:
+(
+x
++
+y
++
+z
+)
+≤
+(
+x
++
+y
++
+z
+)
+(
++
+y
++
+z
+)
+⇒
+x
+x
++
+y
++
+z
+≤
+x
+(
++
+y
++
+z
+)
+(
+x
++
+y
++
+z
+)
+.
+(x
+ +y
+ +z
+ )
+ ≤(x
+ +y
+ +z
+ )(1+y
+ +z
+ )⇒
+x
+ +y
+ +z
+x
+ ≤
+(x
+ +y
+ +z
+ )
+x(1+y
+ +z
+ )
+ .
+Hence, the right-hand side (RHS) satisfies
+\begin{align*}
+\text{RHS} &\leq \frac{x(1 + y^2 + z^2) + y(1 + z^2 + x^2) + z(1 + x^2 + y^2)}{(x^2 + y^2 + z^2)^2} \
+&= \frac{x+y+z + (xy+yz+zx)(x+y+z) - 3xyz}{(x^2 + y^2 + z^2)^2}.
+\end{align*}
+Since <math>x^2+y^2+z^2 \geq \frac{t^2}{3}</math> and <math>xy+yz+zx \leq \frac{t^2}{3}</math>, we have
+RHS
+≤
+t
++
+t
+t
+/
+=
+t
+(
+t
++
+t
+)
+=
+t
+⋅
+t
++
+t
+=
+(
+t
++
+t
+)
+t
+=
+t
++
+t
+t
+.
+RHS≤
+t
+ /9
+t+
+t
+ =
+t
+ (t+
+t
+ )=
+t
+ ⋅
+t+t
+ =
+t
+(t
+ +3t)
+ =
+t
+t
+ +9t
+ .
+Wait, the original says:
+≤
+t
++
+t
+t
+=
+t
++
+t
+−
+t
+≤
+⇔
+(
+t
+−
+)
+(
+t
+−
+)
+≥
+≤
+t
+t+
+t
+ =
+t
+t
+ +9t−27
+ ≤1⇔(t−3)(t
+ −9)≥0
+But there is a discrepancy: The step from the previous expression to the next is not clear. Actually, the original says:
+≤
+t
++
+t
+t
+=
+t
++
+t
+−
+t
+≤
+t
+t+
+t
+ =
+t
+t
+ +9t−27
+That seems to have an error: Actually, the original might have intended:
+We have:
+x
++
+y
++
+z
++
+(
+x
+y
++
+y
+z
++
+z
+x
+)
+(
+x
++
+y
++
+z
+)
+−
+x
+y
+z
+(
+x
++
+y
++
+z
+)
+≤
+t
++
+(
+t
+/
+)
+t
+t
+/
+=
+t
++
+t
+/
+t
+/
+=
+t
++
+t
+/
+t
+=
+t
++
+t
+t
+=
+t
++
+t
+t
+.
+(x
+ +y
+ +z
+ )
+x+y+z+(xy+yz+zx)(x+y+z)−3xyz
+ ≤
+t
+ /9
+t+(t
+ /3)t
+ =
+t
+ /9
+t+t
+ /3
+ =9
+t
+t+t
+ /3
+ =
+t
+t+3t
+ =
+t
+t
+ +9t
+ .
+But then the original writes:
+≤
+t
++
+t
+t
+=
+t
++
+t
+−
+t
+≤
+t
+t+
+t
+ =
+t
+t
+ +9t−27
+That is inconsistent. Possibly there is a misprint. Actually, the original then says:
+t
++
+t
+−
+t
+≤
+⇔
+(
+t
+−
+)
+(
+t
+−
+)
+≥
+.
+t
+t
+ +9t−27
+ ≤1⇔(t−3)(t
+ −9)≥0.
+But if we had <math>\frac{3t^3+9t}{t^4}</math>, then we would need to show that <math>\frac{3t^3+9t}{t^4} \leq 1</math> for <math>t \geq 3</math>, which is <math>3t^3+9t \leq t^4</math> or <math>t^4 - 3t^3 - 9t \geq 0</math>, i.e., <math>t(t^3-3t^2-9) \geq 0</math>. That is not the same as <math>(t-3)(t^3-9) \geq 0</math>. So there is a discrepancy.
+Let's re-read the original:
+"RHS <math>\leq \frac{x+y+z+(xy+yz+zx)(x+y+z)-3xyz}{(x^2 + y^2 + z^2)^2}
+\leq \frac{t + \frac{t^3}{3}}{t^4} = \frac{3t^3 + 9t - 27}{t^4} \leq 1</math>"
+Wait, the original has:
+≤
+t
++
+t
+t
+=
+t
++
+t
+−
+t
+≤
+t
+t+
+t
+ =
+t
+t
+ +9t−27
+But <math>\frac{t + \frac{t^3}{3}}{t^4} = \frac{3t + t^3}{3t^4} = \frac{t^3+3t}{3t^4}</math>, not <math>\frac{3t^3+9t-27}{t^4}</math>. So there is a mistake.
+Maybe it is:
+≤
+t
++
+t
+−
+(
+x
++
+y
++
+z
+)
+≤
+t
++
+t
+−
+t
+/
+=
+(
+t
++
+t
+/
+−
+)
+t
+=
+t
++
+t
+−
+t
+=
+t
++
+t
+−
+t
+≤
+(x
+ +y
+ +z
+ )
+t+
+t
+ −3
+ ≤
+t
+ /9
+t+
+t
+ −3
+ =
+t
+(t+t
+ /3−3)
+ =
+t
+t+3t
+ −27
+ =
+t
+t
+ +9t−27
+because we also subtract 3xyz? Actually, note that the numerator is:
+x+y+z + (xy+yz+zx)(x+y+z) - 3xyz.
+And we have xyz >= 1, so -3xyz <= -3. So we can bound:
+x+y+z + (xy+yz+zx)(x+y+z) - 3xyz <= t + (t^2/3)*t - 3 = t + t^3/3 - 3.
+And then denominator: (x^2+y^2+z^2)^2 >= (t^2/3)^2 = t^4/9.
+So indeed:
+RHS <= (t + t^3/3 - 3) / (t^4/9) = 9(t + t^3/3 - 3)/t^4 = (9t + 3t^3 - 27)/t^4 = (3t^3+9t-27)/t^4.
+So that step is correct:
+RHS
+≤
+t
++
+t
+−
+t
+.
+RHS≤
+t
+t
+ +9t−27
+ .
+Then we want to show that this is ≤ 1, i.e.,
+t
++
+t
+−
+t
+≤
+⇔
+t
++
+t
+−
+≤
+t
+⇔
+t
+−
+t
+−
+t
++
+≥
+.
+t
+t
+ +9t−27
+ ≤1⇔3t
+ +9t−27≤t
+ ⇔t
+ −3t
+ −9t+27≥0.
+Factor: t^4 - 3t^3 - 9t + 27 = (t-3)(t^3 - 9) ? Let's check: (t-3)(t^3 - 9) = t^4 - 9t - 3t^3 + 27 = t^4 - 3t^3 - 9t + 27. Yes.
+So we need (t-3)(t^3-9) >= 0. For t>=3, t-3>=0 and t^3-9>=0, so indeed it holds. So RHS ≤ 1.
+Then the original says: "By AM-GM: LHS ≥ ...". But careful: The LHS here is the left-hand side of the inequality we want to prove? Actually, we want to prove that LHS (which is the sum of fractions with x^4) is at least 1. So we need to show LHS ≥ 1.
+The original then says:
+LHS ≥ (x^3+y^3+z^3)^2 / (x^6+y^6+z^6+2(x^2y^2+y^2z^2+z^2x^2)) ≥ 1.
+That is by Cauchy-Schwarz? Actually, by Titu's lemma (Cauchy-Schwarz in Engel form):
+∑ x^4/(x^4+y^2+z^2) ≥ (∑ x^2)^2 / ∑ (x^4+y^2+z^2) = (x^2+y^2+z^2)^2 / (∑ x^4 + 2∑ x^2) but careful: Actually, the denominators are different: The first term denominator is x^4+y^2+z^2, so if we sum over cyclic, then the denominator sum becomes (x^4+y^4+z^4) + (y^2+z^2) + (z^2+x^2) + (x^2+y^2) = x^4+y^4+z^4 + 2(x^2+y^2+z^2). So then by Cauchy, we have:
+LHS = ∑ x^4/(x^4+y^2+z^2) ≥ (x^2+y^2+z^2)^2 / (x^4+y^4+z^4+2(x^2+y^2+z^2)).
+But the original writes:
+LHS ≥ (x^3+y^3+z^3)^2 / (x^6+y^6+z^6+2(x^2y^2+y^2z^2+z^2x^2)).
+That is different: They have x^3 in numerator squared, and denominator: x^6+y^6+z^6+2(x^2y^2+...). Actually, note that (x^3)^2 = x^6. So it is similar but with squares? Actually, if we set a = x^2, then a^2 = x^4. So it's not exactly.
+Wait, maybe they used:
+By Cauchy, (∑ x^4/(x^4+y^2+z^2)) (∑ x^4(x^4+y^2+z^2)) ≥ (∑ x^4)^2.
+But then ∑ x^4(x^4+y^2+z^2) = ∑ (x^8 + x^4y^2 + x^4z^2) = ∑ x^8 + ∑ x^4y^2 + ∑ x^4z^2.
+That is not the same as x^6+y^6+z^6+2(x^2y^2+...)? Actually, x^6? No.
+Alternatively, they might have used the inequality:
+∑ x^4/(x^4+y^2+z^2) ≥ (∑ x^2)^2 / (∑ x^4 + 2∑ x^2)
+but then they claim that is ≥ 1 if (x^2+y^2+z^2)^2 ≥ x^4+y^4+z^4+2(x^2+y^2+z^2) which is not necessarily true.
+Maybe they used a different approach: They said:
+LHS ≥ (x^3+y^3+z^3)^2 / (x^6+y^6+z^6+2(x^2y^2+y^2z^2+z^2x^2)).
+That would come from Cauchy if we write:
+(∑ x^4/(x^4+y^2+z^2)) (∑ (x^4+y^2+z^2)) ≥ (∑ x^2)^2? Not exactly.
+Wait, consider:
+We want to lower bound LHS = ∑ x^4/(x^4+y^2+z^2).
+By Cauchy, (∑ x^4/(x^4+y^2+z^2)) (∑ x^4(x^4+y^2+z^2)) ≥ (∑ x^4)^2.
+But then that gives LHS ≥ (∑ x^4)^2 / (∑ (x^8 + x^4y^2+x^4z^2)) = (∑ x^4)^2 / (∑ x^8 + ∑ x^4y^2+∑ x^4z^2).
+And note that ∑ x^8 = x^8+y^8+z^8, and ∑ x^4y^2+∑ x^4z^2 = ∑ x^4(y^2+z^2) = ∑ x^4((x^2+y^2+z^2)-x^2) = (x^2+y^2+z^2)∑ x^4 - ∑ x^6.
+So then denominator = ∑ x^8 + (x^2+y^2+z^2)∑ x^4 - ∑ x^6.
+That is not the same as x^6+y^6+z^6+2(x^2y^2+y^2z^2+z^2x^2) unless further conditions.
+Alternatively, maybe they used:
+By Titu's lemma:
+∑ x^4/(x^4+y^2+z^2) ≥ (∑ x^2)^2 / (∑ (x^4+y^2+z^2)) = (x^2+y^2+z^2)^2 / (∑ x^4 + 2∑ x^2).
+But then they claim that is ≥ 1 if (x^2+y^2+z^2)^2 ≥ ∑ x^4+2∑ x^2, i.e., 2(x^2y^2+y^2z^2+z^2x^2) ≥ 2(x^2+y^2+z^2) or x^2y^2+y^2z^2+z^2x^2 ≥ x^2+y^2+z^2, which is not necessarily true.
+Given that the original text says:
+"LHS ≥ \frac{(x^3 + y^3 + z^3)^2}{x^6 + y^6 + z^6 + 2(x^2y^2 + y^2z^2 + z^2x^2)} \geq 1"
+Maybe it is actually:
+LHS = ∑ x^4/(x^4+y^2+z^2) and they use the substitution: Let a = x^2, b = y^2, c = z^2. Then LHS = ∑ a^2/(a^2+b+c). And then by Cauchy,
+(∑ a^2/(a^2+b+c)) (∑ a^2(a^2+b+c)) ≥ (∑ a^2)^2.
+But ∑ a^2(a^2+b+c) = ∑ (a^4 + a^2b + a^2c) = ∑ a^4 + ∑ a^2(b+c) = ∑ a^4 + ∑ a^2((a+b+c)-a) = ∑ a^4 + (a+b+c)∑ a^2 - ∑ a^3.
+So then LHS ≥ (∑ a^2)^2 / (∑ a^4 + (a+b+c)∑ a^2 - ∑ a^3). That doesn't simplify to the given.
+Alternatively, they might have used the inequality:
+a^2/(a^2+b+c) ≥ (a^3)^2/(a^6 + a^3(b+c))? That is not standard.
+Wait, the given expression:
+\frac{(x^3+y^3+z^3)^2}{x^6+y^6+z^6+2(x^2y^2+y^2z^2+z^2x^2)}
+Notice that x^6+y^6+z^6+2(x^2y^2+y^2z^2+z^2x^2) = (x^2+y^2+z^2)^3? Actually, (x^2+y^2+z^2)^3 = x^6+y^6+z^6+3(x^2y^2(x^2+y^2)+... so no.
+Maybe it's a misprint: They might have meant:
+LHS ≥ \frac{(x^2+y^2+z^2)^2}{x^4+y^4+z^4+2(x^2+y^2+z^2)}.
+But then they claim that is ≥ 1 if (x^2+y^2+z^2)^2 ≥ x^4+y^4+z^4+2(x^2+y^2+z^2) which is equivalent to 2(x^2y^2+y^2z^2+z^2x^2) ≥ 2(x^2+y^2+z^2) or x^2y^2+y^2z^2+z^2x^2 ≥ x^2+y^2+z^2. That is not generally true.
+Given the subsequent steps:
+They then say:
+≥ 1 ⇔ x^3y^3+y^3z^3+z^3x^3 ≥ x^2y^2+y^2z^2+z^2x^2 ⇔ a^3+b^3+c^3 ≥ a^2+b^2+c^2, where a=xy, b=yz, c=zx.
+So indeed, they are comparing:
+(x^3+y^3+z^3)^2 and x^6+y^6+z^6+2(x^2y^2+y^2z^2+z^2x^2).
+Note that (x^3+y^3+z^3)^2 = x^6+y^6+z^6+2(x^3y^3+x^3z^3+y^3z^3).
+So the inequality
+\frac{(x^3+y^3+z^3)^2}{x^6+y^6+z^6+2(x^2y^2+y^2z^2+z^2x^2)} \geq 1
+is equivalent to
+x^6+y^6+z^6+2(x^3y^3+x^3z^3+y^3z^3) \geq x^6+y^6+z^6+2(x^2y^2+y^2z^2+z^2x^2)
+i.e., x^3y^3+x^3z^3+y^3z^3 \geq x^2y^2+y^2z^2+z^2x^2.
+So that step is correct: They claim that LHS (the sum of fractions) is at least that fraction, and then that fraction is at least 1 if x^3y^3+... >= x^2y^2+... So then they set a=xy, b=yz, c=zx, so that a^3+b^3+c^3 >= a^2+b^2+c^2.
+And then they note that abc = x^2y^2z^2, and since x,y,z are positive with xyz>=1, then abc>=1.
+And then they use Holder: (a^2+b^2+c^2)^3 <= 3(a^3+b^3+c^3)^2 <= (a^3+b^3+c^3)^3? Actually, they write:
+By Holder: (a^2+b^2+c^2)^3 <= 3(a^3+b^3+c^3)^2 <= (a^3+b^3+c^3)^3 ⇒ a^2+b^2+c^2 <= a^3+b^3+c^3.
+Wait, check: Holder says: (a^3+b^3+c^3)(1+1+1)(1+1+1) >= (a+b+c)^3, but that's not directly.
+They claim: (a^2+b^2+c^2)^3 <= 3(a^3+b^3+c^3)^2. Is that true?
+By power mean, ( (a^3+b^3+c^3)/3 )^(1/3) >= ( (a^2+b^2+c^2)/3 )^(1/2) so (a^3+b^3+c^3) >= 3^{-1/2} (a^2+b^2+c^2)^(3/2)? That gives (a^3+b^3+c^3)^2 >= (1/3)(a^2+b^2+c^2)^3, so actually (a^2+b^2+c^2)^3 <= 3(a^3+b^3+c^3)^2. So that is true.
+Then they claim: 3(a^3+b^3+c^3)^2 <= (a^3+b^3+c^3)^3, which would require a^3+b^3+c^3 >= 3. And since a,b,c are such that abc>=1, by AM-GM, a^3+b^3+c^3 >= 3(abc) >= 3. So indeed, a^3+b^3+c^3 >= 3. Then we have:
+(a^2+b^2+c^2)^3 <= 3(a^3+b^3+c^3)^2 <= (a^3+b^3+c^3)^3, so taking cube roots: a^2+b^2+c^2 <= a^3+b^3+c^3.
+So that step is valid.
+But then the claim that LHS >= that fraction is not justified in the text. They simply say "By AM-GM:" and then write that inequality. Possibly they meant by Cauchy-Schwarz? Let's check: We want to show:
+∑ x^4/(x^4+y^2+z^2) >= (x^3+y^3+z^3)^2 / (x^6+y^6+z^6+2(x^2y^2+y^2z^2+z^2x^2)).
+This is equivalent to:
+(∑ x^4/(x^4+y^2+z^2)) (x^6+y^6+z^6+2(x^2y^2+y^2z^2+z^2x^2)) >= (x^3+y^3+z^3)^2.
+Now, by Cauchy, we have:
+(∑ x^4/(x^4+y^2+z^2)) (∑ x^4(x^4+y^2+z^2)) >= (∑ x^4)^2.
+But we need (∑ x^3)^2 on the right. So if we can show that (∑ x^4)^2 >= (∑ x^3)^2, i.e., ∑ x^4 >= ∑ x^3, that is not always true. Alternatively, if we use the weights differently:
+We want to relate x^4(x^4+y^2+z^2) to something like x^6? Not sure.
+Maybe they used the following:
+x^4/(x^4+y^2+z^2) >= x^6/(x^6+x^2y^2+x^2z^2) because multiply numerator and denominator by x^2 gives: x^6/(x^6+x^2y^2+x^2z^2). And then sum cyclically:
+LHS >= ∑ x^6/(x^6+x^2y^2+x^2z^2).
+Then by Cauchy,
+∑ x^6/(x^6+x^2y^2+x^2z^2) >= (∑ x^3)^2 / (∑ x^6+∑ x^2y^2+∑ x^2z^2) = (∑ x^3)^2 / (∑ x^6+2∑ x^2y^2).
+That is exactly the claimed lower bound. And the inequality
+x^4/(x^4+y^2+z^2) >= x^6/(x^6+x^2y^2+x^2z^2)
+is equivalent to cross multiply: x^4(x^6+x^2y^2+x^2z^2) >= x^6(x^4+y^2+z^2) => x^10+x^6y^2+x^6z^2 >= x^10+x^6y^2+x^6z^2, which is equality! Actually, check:
+x^4/(x^4+y^2+z^2) >= x^6/(x^6+x^2y^2+x^2z^2)
+Multiply both sides by positive denominators:
+x^4(x^6+x^2y^2+x^2z^2) >= x^6(x^4+y^2+z^2)
+=> x^10 + x^6y^2 + x^6z^2 >= x^10 + x^6y^2 + x^6z^2.
+So it's actually equality. So indeed,
+x^4/(x^4+y^2+z^2) = x^6/(x^6+x^2y^2+x^2z^2).
+Because multiply numerator and denominator by x^2 gives that provided x>0. So we have:
+LHS = ∑ x^4/(x^4+y^2+z^2) = ∑ x^6/(x^6+x^2y^2+x^2z^2).
+Then by Cauchy-Schwarz,
+∑ x^6/(x^6+x^2y^2+x^2z^2) ≥ (∑ x^3)^2 / (∑ x^6 + ∑ x^2y^2+∑ x^2z^2) = (∑ x^3)^2 / (∑ x^6+2∑ x^2y^2).
+So that step is justified.
+So we will include that step.
+Now, let's write the LaTeX code accordingly.
+I'll structure it as:
+\documentclass{article}
+\usepackage{amsmath, amssymb}
+\begin{document}
+\section*{Solution}
+First, we can easily see that
+x
+−
+x
+x
++
+y
++
+z
+≥
+x
+−
+x
+x
++
+y
++
+z
+x
+ +y
+ +z
+x
+ −x
+ ≥
+x
+ +y
+ +z
+x
+ −x
+since
+x
+−
+x
+x
++
+y
++
+z
+−
+x
+−
+x
+x
++
+y
++
+z
+=
+x
+(
+x
+−
+)
+(
+x
++
+x
++
+)
+(
+y
++
+z
+)
+(
+x
++
+y
++
+z
+)
+(
+x
++
+y
++
+z
+)
+≥
+,
+x
+ +y
+ +z
+x
+ −x
+ −
+x
+ +y
+ +z
+x
+ −x
+ =
+(x
+ +y
+ +z
+ )(x
+ +y
+ +z
+ )
+x(x−1)
+ (x
+ +x+1)(y
+ +z
+ )
+ ≥0,
+which holds for all positive real numbers <math>x, y, z</math>.
+Thus, it is enough to prove that
+x
+−
+x
+x
++
+y
++
+z
++
+y
+−
+y
+y
++
+z
++
+x
++
+z
+−
+z
+z
++
+x
++
+y
+≥
+.
+x
+ +y
+ +z
+x
+ −x
+ +
+y
+ +z
+ +x
+y
+ −y
+ +
+z
+ +x
+ +y
+z
+ −z
+ ≥0.
+Rewriting, we need to show
+x
+x
++
+y
++
+z
++
+y
+y
++
+z
++
+x
++
+z
+z
++
+x
++
+y
+≥
+x
+x
++
+y
++
+z
++
+y
+y
++
+z
++
+x
++
+z
+z
++
+x
++
+y
+.
+x
+ +y
+ +z
+x
+ +
+y
+ +z
+ +x
+y
+ +
+z
+ +x
+ +y
+z
+ ≥
+x
+ +y
+ +z
+x
+ +
+y
+ +z
+ +x
+y
+ +
+z
+ +x
+ +y
+z
+ .
+Let <math>t = x + y + z</math>. By the AM–GM inequality, we have <math>t \geq 3\sqrt[3]{xyz} \geq 3</math>, and also <math>xy + yz + zx \leq \frac{t^2}{3} \leq x^2 + y^2 + z^2</math>.
+Now, estimate the right-hand side (RHS). By the Cauchy–Schwarz inequality,
+(
+x
++
+y
++
+z
+)
+≤
+(
+x
++
+y
++
+z
+)
+(
++
+y
++
+z
+)
+,
+(x
+ +y
+ +z
+ )
+ ≤(x
+ +y
+ +z
+ )(1+y
+ +z
+ ),
+so
+x
+x
++
+y
++
+z
+≤
+x
+(
++
+y
++
+z
+)
+(
+x
++
+y
++
+z
+)
+.
+x
+ +y
+ +z
+x
+ ≤
+(x
+ +y
+ +z
+ )
+x(1+y
+ +z
+ )
+ .
+Cyclically summing, we get
+\begin{align*}
+\text{RHS} &\leq \frac{x(1+y^2+z^2) + y(1+z^2+x^2) + z(1+x^2+y^2)}{(x^2+y^2+z^2)^2} \
+&= \frac{x+y+z + (xy+yz+zx)(x+y+z) - 3xyz}{(x^2+y^2+z^2)^2}.
+\end{align*}
+Since <math>xyz \geq 1</math>, we have <math>-3xyz \leq -3</math>. Also, <math>xy+yz+zx \leq \frac{t^2}{3}</math> and <math>x^2+y^2+z^2 \geq \frac{t^2}{3}</math>. Hence,
+\begin{align*}
+\text{RHS} &\leq \frac{t + \frac{t^2}{3} \cdot t - 3}{(t^2/3)^2} = \frac{t + \frac{t^3}{3} - 3}{t^4/9} = \frac{9(t + t^3/3 - 3)}{t^4} = \frac{9t + 3t^3 - 27}{t^4} = \frac{3t^3 + 9t - 27}{t^4}.
+\end{align*}
+We claim that
+t
++
+t
+−
+t
+≤
+.
+t
+t
+ +9t−27
+ ≤1.
+This is equivalent to
+t
++
+t
+−
+≤
+t
+⇔
+t
+−
+t
+−
+t
++
+≥
+.
+t
+ +9t−27≤t
+ ⇔t
+ −3t
+ −9t+27≥0.
+Factoring, we have
+t
+−
+t
+−
+t
++
+=
+(
+t
+−
+)
+(
+t
+−
+)
+≥
+,
+t
+ −3t
+ −9t+27=(t−3)(t
+ −9)≥0,
+which holds for <math>t \geq 3</math>. Therefore, RHS <math>\leq 1</math>.
+Now, estimate the left-hand side (LHS). Note that for <math>x > 0</math>, we have
+x
+x
++
+y
++
+z
+=
+x
+x
++
+x
+y
++
+x
+z
+.
+x
+ +y
+ +z
+x
+ =
+x
+ +x
+ y
+ +x
+ z
+x
+ .
+Thus,
+LHS
+=
+∑
+x
+x
++
+y
++
+z
+=
+∑
+x
+x
++
+x
+y
++
+x
+z
+.
+LHS=∑
+x
+ +y
+ +z
+x
+ =∑
+x
+ +x
+ y
+ +x
+ z
+x
+ .
+By the Cauchy–Schwarz inequality,
+(
+∑
+x
+x
++
+x
+y
++
+x
+z
+)
+(
+∑
+(
+x
++
+x
+y
++
+x
+z
+)
+)
+≥
+(
+∑
+x
+)
+.
+(∑
+x
+ +x
+ y
+ +x
+ z
+x
+ )(∑(x
+ +x
+ y
+ +x
+ z
+ ))≥(∑x
+ )
+ .
+But
+∑
+(
+x
++
+x
+y
++
+x
+z
+)
+=
+∑
+x
++
+∑
+x
+y
++
+∑
+x
+z
+=
+∑
+x
++
+∑
+x
+y
+.
+∑(x
+ +x
+ y
+ +x
+ z
+ )=∑x
+ +∑x
+ y
+ +∑x
+ z
+ =∑x
+ +2∑x
+ y
+ .
+Hence,
+LHS
+≥
+(
+x
++
+y
++
+z
+)
+x
++
+y
++
+z
++
+(
+x
+y
++
+y
+z
++
+z
+x
+)
+.
+LHS≥
+x
+ +y
+ +z
+ +2(x
+ y
+ +y
+ z
+ +z
+ x
+ )
+(x
+ +y
+ +z
+ )
+ .
+We now show that this expression is at least 1. That is, we need
+(
+x
++
+y
++
+z
+)
+≥
+x
++
+y
++
+z
++
+(
+x
+y
++
+y
+z
++
+z
+x
+)
+.
+(x
+ +y
+ +z
+ )
+ ≥x
+ +y
+ +z
+ +2(x
+ y
+ +y
+ z
+ +z
+ x
+ ).
+Expanding the left-hand side:
+(
+x
++
+y
++
+z
+)
+=
+x
++
+y
++
+z
++
+(
+x
+y
++
+x
+z
++
+y
+z
+)
+.
+(x
+ +y
+ +z
+ )
+ =x
+ +y
+ +z
+ +2(x
+ y
+ +x
+ z
+ +y
+ z
+ ).
+So the inequality becomes
+x
+y
++
+x
+z
++
+y
+z
+≥
+x
+y
++
+y
+z
++
+z
+x
+.
+x
+ y
+ +x
+ z
+ +y
+ z
+ ≥x
+ y
+ +y
+ z
+ +z
+ x
+ .
+Let <math>a = xy</math>, <math>b = yz</math>, <math>c = zx</math>. Then the inequality is equivalent to
+a
++
+b
++
+c
+≥
+a
++
+b
++
+c
+.
+a
+ +b
+ +c
+ ≥a
+ +b
+ +c
+ .
+Note that <math>abc = x^2y^2z^2 \geq 1</math> since <math>xyz \geq 1</math>. Also, by the AM–GM inequality, <math>a^3 + b^3 + c^3 \geq 3\sqrt[3]{a^3b^3c^3} = 3abc \geq 3</math>.
+Now, by the power mean inequality, we have
+(
+a
++
+b
++
+c
+)
+/
+≥
+(
+a
++
+b
++
+c
+)
+/
+,
+(
+a
+ +b
+ +c
+ )
+/3
+ ≥(
+a
+ +b
+ +c
+ )
+/2
+ ,
+which implies
+(
+a
++
+b
++
+c
+)
+≥
+(
+a
++
+b
++
+c
+)
+.
+(a
+ +b
+ +c
+ )
+ ≥
+ (a
+ +b
+ +c
+ )
+ .
+That is,
+(
+a
++
+b
++
+c
+)
+≤
+(
+a
++
+b
++
+c
+)
+.
+(a
+ +b
+ +c
+ )
+ ≤3(a
+ +b
+ +c
+ )
+ .
+Since <math>a^3 + b^3 + c^3 \geq 3</math>, we have
+(
+a
++
+b
++
+c
+)
+≤
+(
+a
++
+b
++
+c
+)
+.
+(a
+ +b
+ +c
+ )
+ ≤(a
+ +b
+ +c
+ )
+ .
+Combining, we get
+(
+a
++
+b
++
+c
+)
+≤
+(
+a
++
+b
++
+c
+)
+,
+(a
+ +b
+ +c
+ )
+ ≤(a
+ +b
+ +c
+ )
+ ,
+so taking cube roots yields
+a
++
+b
++
+c
+≤
+a
++
+b
++
+c
+,
+a
+ +b
+ +c
+ ≤a
+ +b
+ +c
+ ,
+as desired. Therefore, LHS <math>\geq 1</math>.
+Since we have shown that LHS <math>\geq 1</math> and RHS <math>\leq 1</math>, it follows that
+x
+x
++
+y
++
+z
++
+y
+y
++
+z
++
+x
++
+z
+z
++
+x
++
+y
+≥
+x
+x
++
+y
++
+z
++
+y
+y
++
+z
++
+x
++
+z
+z
++
+x
++
+y
+,
+x
+ +y
+ +z
+x
+ +
+y
+ +z
+ +x
+y
+ +
+z
+ +x
+ +y
+z
+ ≥
+x
+ +y
+ +z
+x
+ +
+y
+ +z
+ +x
+y
+ +
+z
+ +x
+ +y
+z
+ ,
+and hence the original inequality holds.
+Equality occurs when <math>x = y = z = 1</math>.
+\end{document}