2005 IMO Problems/Problem 3

""" 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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2005 IMO Problems/Problem 3