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

Latest revision as of 03:37, 4 September 2025

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

\section*{Problem} Let x , y , z x,y,z be positive real numbers such that x y z ≥ 1 xyz≥1. Prove that:

x 5 − x 2 x 5 + y 2 + z 2 + y 5 − y 2 y 5 + z 2 + x 2 + z 5 − z 2 z 5 + x 2 + y 2 ≥ 0. x 5

+y

2

+z

2

x 5

−x

2

y 5

+z

2

+x

2

y 5

−y

2

z 5

+x

2

+y

2

z 5

−z

2

≥0.

\section*{Solution} We first show that for any positive real numbers x , y , z x,y,z, we have

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

Indeed, consider the difference: \begin{align*} &\frac{x^5 - x^2}{x^5 + y^2 + z^2} - \frac{x^4 - x}{x^4 + y^2 + z^2} \ &= \frac{(x^5 - x^2)(x^4 + y^2 + z^2) - (x^4 - x)(x^5 + y^2 + z^2)}{(x^5 + y^2 + z^2)(x^4 + y^2 + z^2)}. \end{align*} The numerator simplifies as follows: \begin{align*} &(x^5 - x^2)(x^4 + y^2 + z^2) - (x^4 - x)(x^5 + y^2 + z^2) \ &= x^5x^4 + x^5(y^2+z^2) - x^2x^4 - x^2(y^2+z^2) - x^4x^5 - x^4(y^2+z^2) + x x^5 + x(y^2+z^2) \ &= x^5(y^2+z^2) - x^2(y^2+z^2) - x^4(y^2+z^2) + x(y^2+z^2) \ &= (y^2+z^2)(x^5 - x^4 - x^2 + x) \ &= (y^2+z^2)x(x^4 - x^3 - x + 1) \quad \text{(but wait, check: actually, } x^5 - x^4 - x^2 + x = x(x^4 - x^3 - x + 1) \text{? That doesn't factor nicely)}. \end{align*} Alternatively, note that:

x 5 − x 2 = x 2 ( x 3 − 1 ) = x 2 ( x − 1 ) ( x 2 + x + 1 ) , x 4 − x = x ( x 3 − 1 ) = x ( x − 1 ) ( x 2 + x + 1 ) . x 5

−x

2

=x

2

(x

3

−1)=x

2

(x−1)(x

2

+x+1),x

4

−x=x(x

3

−1)=x(x−1)(x

2

+x+1).

So then: \begin{align*} &\frac{x^5 - x^2}{x^5 + y^2 + z^2} - \frac{x^4 - x}{x^4 + y^2 + z^2} \ &= \frac{x^2(x-1)(x^2+x+1)}{x^5 + y^2 + z^2} - \frac{x(x-1)(x^2+x+1)}{x^4 + y^2 + z^2} \ &= x(x-1)(x^2+x+1) \left( \frac{x}{x^5 + y^2 + z^2} - \frac{1}{x^4 + y^2 + z^2} \right). \end{align*} Now, combine the terms inside the parentheses:

x x 5 + y 2 + z 2 − 1 x 4 + y 2 + z 2 = x ( x 4 + y 2 + z 2 ) − ( x 5 + y 2 + z 2 ) ( x 5 + y 2 + z 2 ) ( x 4 + y 2 + z 2 ) = ( x 5 + x ( y 2 + z 2 ) ) − ( x 5 + y 2 + z 2 ) ( x 5 + y 2 + z 2 ) ( x 4 + y 2 + z 2 ) = ( x − 1 ) ( y 2 + z 2 ) ( x 5 + y 2 + z 2 ) ( x 4 + y 2 + z 2 ) . x 5

+y

2

+z

2

x

−

x 4

+y

2

+z

2

1

(x 5

+y

2

+z

2

)(x

4

+y

2

+z

2

x(x 4

+y

2

+z

2

)−(x

5

+y

2

+z

2

(x 5

+y

2

+z

2

)(x

4

+y

2

+z

2

(x 5

+x(y

2

+z

2

))−(x

5

+y

2

+z

2

(x 5

+y

2

+z

2

)(x

4

+y

2

+z

2

(x−1)(y 2

+z

2

Thus, the entire difference becomes:

x ( x − 1 ) ( x 2 + x + 1 ) ⋅ ( x − 1 ) ( y 2 + z 2 ) ( x 5 + y 2 + z 2 ) ( 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(x−1)(x 2

+x+1)⋅

(x 5

+y

2

+z

2

)(x

4

+y

2

+z

2

(x−1)(y 2

+z

2

(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.

Hence,

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

Similarly, we have the analogous inequalities for y y and z z. Therefore, it suffices 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,

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 . (1) 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

.(1)

Let t = x + y + z t=x+y+z. Since x , y , z > 0 x,y,z>0 and x y z ≥ 1 xyz≥1, by AM–GM we have t ≥ 3 x y z 3 ≥ 3 t≥3 3

xyz

≥3. Also, note that

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

and

x 2 + y 2 + z 2 ≥ t 2 3 x 2

+y

2

+z

2

≥

3 t 2

Now, we estimate the right-hand side of (1). By the Cauchy–Schwarz inequality, we have:

( 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

),

since

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

+y

2

+z

2

=(x

2

⋅1+y⋅y+z⋅z)

2

≤(x

4

+y

2

+z

2

)(1

2

+y

2

+z

2

)=(x

4

+y

2

+z

2

)(1+y

2

+z

2

).

It follows that

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

+y

2

+z

2

1

≤

(x 2

+y

2

+z

2

1+y 2

+z

2

and multiplying by x x (which is positive) gives:

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

Similarly,

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

+z

2

+x

2

y

≤

(x 2

+y

2

+z

2

y(1+z 2

+x

2

z 4

+x

2

+y

2

z

≤

(x 2

+y

2

+z

2

z(1+x 2

+y

2

Summing these, we obtain:

∑ x x 4 + y 2 + z 2 ≤ 1 ( x 2 + y 2 + z 2 ) 2 [ ∑ x + ∑ x ( y 2 + z 2 ) ] . ∑ x 4

+y

2

+z

2

x

≤

(x 2

+y

2

+z

2

1

[∑x+∑x(y

2

+z

2

)].

Now, note that

∑ x ( y 2 + z 2 ) = ∑ ( x y 2 + x z 2 ) = ( x y 2 + x z 2 ) + ( y z 2 + y x 2 ) + ( z x 2 + z y 2 ) = ( x + y + z ) ( x y + y z + z x ) − 3 x y z . ∑x(y 2

+z

2

)=∑(xy

2

+xz

2

)=(xy

2

+xz

2

)+(yz

2

+yx

2

)+(zx

2

+zy

2

)=(x+y+z)(xy+yz+zx)−3xyz.

Thus,

∑ x x 4 + y 2 + z 2 ≤ t + t ( x y + y z + z x ) − 3 x y z ( x 2 + y 2 + z 2 ) 2 . ∑ x 4

+y

2

+z

2

x

≤

(x 2

+y

2

+z

2

t+t(xy+yz+zx)−3xyz

Since x y z ≥ 1 xyz≥1, we have − 3 x y z ≤ − 3 −3xyz≤−3. Also, x y + y z + z x ≤ t 2 3 xy+yz+zx≤ 3 t 2

and

x 2 + y 2 + z 2 ≥ t 2 3 x 2

+y

2

+z

2

≥

3 t 2

, so

( x 2 + y 2 + z 2 ) 2 ≥ ( t 2 3 ) 2 = t 4 9 . (x 2

+y

2

+z

2

≥(

3 t 2

2

9 t 4

Therefore,

∑ x x 4 + y 2 + z 2 ≤ t + t ⋅ t 2 3 − 3 ( t 2 / 3 ) 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 4

+y

2

+z

2

x

≤

(t 2

/3)

2

t+t⋅ 3 t 2

−3

t 4

/9

t+ 3 t 3

−3

t 4

9(t+ 3 t 3

−3)

t 4

9t+3t 3

−27

t 4

3t 3

+9t−27

We now show 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.

Factor the left-hand side:

t 4 − 3 t 3 − 9 t + 27 = ( t − 3 ) ( t 3 − 9 ) . t 4

−3t

3

−9t+27=(t−3)(t

3

−9).

For t ≥ 3 t≥3, we have t − 3 ≥ 0 t−3≥0 and t 3 − 9 ≥ 18 t 3

−9≥18, so indeed

( t − 3 ) ( t 3 − 9 ) ≥ 0 (t−3)(t 3

−9)≥0. Hence,

∑ x x 4 + y 2 + z 2 ≤ 1. (2) ∑ x 4

+y

2

+z

2

x

≤1.(2)

Now, we estimate the left-hand side of (1). Note that

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

and similarly for the other terms. So,

∑ 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

By the Cauchy–Schwarz inequality (Titu's lemma), we have:

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

+x

2

+x

2

x 6

≥

∑(x 6

+x

2

+x

2

(x 3

+y

3

+z

3

2

x 6

+y

6

+z

6

+2(x

2

+y

2

+z

2

(x 3

+y

3

+z

3

2

.(3)

We claim that

( 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 + y 3 z 3 + z 3 x 3 ) . (x 3

+y

3

+z

3

2

=x

6

+y

6

+z

6

+2(x

3

+y

3

+z

3

).

Thus, the inequality is equivalent to

x 3 y 3 + y 3 z 3 + z 3 x 3 ≥ x 2 y 2 + y 2 z 2 + z 2 x 2 . (4) x 3

3

+y

3

+z

3

≥x

2

+y

2

+z

2

.(4)

Let a = x y a=xy, b = y z b=yz, c = z x c=zx. Then a b c = x 2 y 2 z 2 ≥ 1 abc=x 2

2

≥1, and inequality (4) becomes:

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

+b

3

+c

3

≥a

2

+b

2

+c

2

We now prove this. By the AM–GM inequality, a 3 + b 3 + c 3 ≥ 3 a b c ≥ 3 a 3

+b

3

+c

3

≥3abc≥3. Also, 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 ≥ 3 ( a 2 + b 2 + c 2 3 ) 3 / 2 = ( a 2 + b 2 + c 2 ) 3 / 2 3 . a 3

+b

3

+c

3

≥3(

3 a 2

+b

2

+c

2

3/2

3

(a 2

+b

2

+c

2

3/2

Squaring both sides (all positive) gives:

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

+b

3

+c

3

2

≥

3 (a 2

+b

2

+c

2

3

That is,

( a 2 + b 2 + c 2 ) 3 ≤ 3 ( a 3 + b 3 + c 3 ) 2 . (5) (a 2

+b

2

+c

2

3

≤3(a

3

+b

3

+c

3

2

.(5)

On the other hand, since a 3 + b 3 + c 3 ≥ 3 a 3

+b

3

+c

3

≥3, we have

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

,because(a

3

+b

3

+c

3

≥3(a

3

+b

3

+c

3

2

⇔a

3

+b

3

+c

3

≥3.

Combining with (5), 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, from (3) we obtain

∑ x 4 x 4 + y 2 + z 2 ≥ 1. (6) ∑ x 4

+y

2

+z

2

x 4

≥1.(6)

Combining (2) and (6), we have

∑ x 4 x 4 + y 2 + z 2 ≥ 1 ≥ ∑ x x 4 + y 2 + z 2 , ∑ x 4

+y

2

+z

2

x 4

≥1≥∑

x 4

+y

2

+z

2

x

which proves (1). Hence, the original inequality holds.

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

\end{document}

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

Latest revision as of 03:37, 4 September 2025

@@ Line 1: / Line 1: @@
-"""
+\documentclass{article}
-Solution:
+\usepackage{amsmath}
+\usepackage{amssymb}
+\begin{document}
+\section*{Problem}
+Let
+x
+,
+y
+,
+z
+x,y,z be positive real numbers such that
+x
+y
+z
+≥
+xyz≥1. Prove that:
-First we can easily see
 x
@@ Line 16: / Line 32: @@
 z
-≥
++
-x
+y
 −
-x
-x
-+
 y
+y
 +
 z
-x
++
+x
++
+z
- +y
+−
+z
-  +z
+z
++
+x
++
+y
+≥
+.
+x
+ +y
+  +z
@@ Line 42: / Line 75: @@
-  ≥
+  +
-x
+y
-  +y
+ +z
+  +x
-  +z
+y
+ −y
-x
- −x
+ +
-⇔
+z
-x
-(
+ +x
-x
-−
+ +y
-)
-(
-x
+z
+ −z
-+
+ ≥0.
+\section*{Solution}
+We first show that for any positive real numbers
 x
-+
+,
-)
-(
 y
+,
-+
 z
+x,y,z, we have
+x
+−
+x
-)
-(
 x
@@ Line 87: / Line 126: @@
 z
-)
+≥
-(
+x
+−
+x
 x
@@ Line 97: / Line 139: @@
 z
-)
+.
-≥
+x
-⇔
-(x
   +y
@@ Line 107: / Line 146: @@
   +z
-  )(x
+x
+ −x
+ ≥
+x
   +y
@@ Line 113: / Line 160: @@
   +z
-  )
-x(x−1)
+x
-  (x
+  −x
- +x+1)(y
- +z
- )
-  ≥0 which is true for
+  .
-x
+Indeed, consider the difference:
-,
+\begin{align*}
-y
+&\frac{x^5 - x^2}{x^5 + y^2 + z^2} - \frac{x^4 - x}{x^4 + y^2 + z^2} \
-,
+&= \frac{(x^5 - x^2)(x^4 + y^2 + z^2) - (x^4 - x)(x^5 + y^2 + z^2)}{(x^5 + y^2 + z^2)(x^4 + y^2 + z^2)}.
-z
+\end{align*}
-x,y,z and
+The numerator simplifies as follows:
-z
+\begin{align*}
-z are positive real numbers.
+&(x^5 - x^2)(x^4 + y^2 + z^2) - (x^4 - x)(x^5 + y^2 + z^2) \
+&= x^5x^4 + x^5(y^2+z^2) - x^2x^4 - x^2(y^2+z^2) - x^4x^5 - x^4(y^2+z^2) + x x^5 + x(y^2+z^2) \
+&= x^5(y^2+z^2) - x^2(y^2+z^2) - x^4(y^2+z^2) + x(y^2+z^2) \
+&= (y^2+z^2)(x^5 - x^4 - x^2 + x) \
+&= (y^2+z^2)x(x^4 - x^3 - x + 1) \quad \text{(but wait, check: actually, } x^5 - x^4 - x^2 + x = x(x^4 - x^3 - x + 1) \text{? That doesn't factor nicely)}.
+\end{align*}
+Alternatively, note that:
-It's enough to prove:
 x
 −
 x
+=
 x
-+
-y
-+
+(
-z
+x
-+
-y
 −
-y
-y
+)
+=
-+
+x
-z
-+
+(
+x
+−
+)
+(
+x
++
+x
++
+)
+,
 x
-+
-z
 −
-z
+x
-z
+=
+x
+(
+x
+−
+)
+=
+x
+(
+x
+−
+)
+(
+x
 +
 x
 +
-y
+)
-≥
+.
 x
-  +y
+  −x
-  +z
+  =x
+  (x
-x
+  −1)=x
-  −x
- +
-y
- +z
-  +x
+  (x−1)(x
+  +x+1),x
-y
-  −y
+  −x=x(x
-  +
+  −1)=x(x−1)(x
-z
- +x
-  +y
+  +x+1).
+So then:
+\begin{align*}
-z
+&\frac{x^5 - x^2}{x^5 + y^2 + z^2} - \frac{x^4 - x}{x^4 + y^2 + z^2} \
+&= \frac{x^2(x-1)(x^2+x+1)}{x^5 + y^2 + z^2} - \frac{x(x-1)(x^2+x+1)}{x^4 + y^2 + z^2} \
- −z
+&= x(x-1)(x^2+x+1) \left( \frac{x}{x^5 + y^2 + z^2} - \frac{1}{x^4 + y^2 + z^2} \right).
+\end{align*}
- ≥0
+Now, combine the terms inside the parentheses:
-Or
 x
 x
 +
 y
@@ Line 223: / Line 276: @@
 z
+−
+x
 +
 y
-y
 +
 z
-+
+=
+x
+(
 x
-+
-z
-z
 +
-x
+y
 +
-y
+z
-≥
+)
+−
+(
 x
-x
 +
 y
@@ Line 255: / Line 308: @@
 z
+)
+(
+x
 +
 y
++
+z
+)
+(
+x
++
 y
 +
 z
+)
+=
+(
+x
 +
 x
+(
+y
 +
 z
-z
+)
-+
+)
+−
+(
 x
 +
 y
-x
++
+z
- +y
- +z
+)
+(
+x
++
+y
++
-x
+z
- +
-y
- +z
- +x
+)
+(
+x
-y
++
+y
- +
-z
- +x
- +y
++
+z
+)
-z
+=
+(
+x
+−
- ≥
-x
+)
+(
- +y
+y
- +z
++
+z
+)
+(
 x
- +
++
 y
- +z
- +x
++
+z
+)
-y
+(
+x
- +
-z
- +x
++
+y
++
+z
+)
+.
+x
   +y
+ +z
-z
+x
+ −
+x
+ +y
+ +z
-Let
-t
+ =
-=
+(x
-x
-+
+  +y
-y
-+
-z
-→
-t
-≥
-(
-x
-y
-z
-)
-/
-≥
-t=x+y+z→t≥3(xyz)
-/3
-  ≥3 and
-x
-y
-+
-y
-z
-+
-z
-x
-≤
-t
+ +z
-≤
-x
-+
+ )(x
-y
+ +y
-+
+ +z
-z
-xy+yz+zx≤
+ )
+x(x
-t
+ +y
+  +z
- ≤x
+ )−(x
   +y
   +z
+  )
-By C-S:
+ =
-(
+(x
-x
+ +y
-+
+ +z
-y
-+
+ )(x
-z
+ +y
-)
+ +z
-≤
+ )
-(
+(x
-x
+ +x(y
-+
-y
+ +z
+ ))−(x
+ +y
-+
+ +z
-z
-)
+ )
-(
+ =
-+
+(x
-y
+ +y
-+
+ +z
-z
-)
+ )(x
-→
-x
-x
-+
+ +y
-y
-+
+ +z
-z
-≤
+ )
+(x−1)(y
+ +z
+ )
+ .
+Thus, the entire difference becomes:
 x
 (
+x
+−
+)
+(
+x
 +
+x
++
+)
+⋅
+(
+x
+−
+)
+(
 y
@@ Line 457: / Line 535: @@
 (
 x
 +
 y
@@ Line 465: / Line 543: @@
 )
+(
+x
++
+y
-(x
++
+z
- +y
+)
+=
+x
+(
+x
+−
+)
- +z
+(
+x
- )
++
+x
++
+)
+(
+y
- ≤(x
++
+z
- +y
- +z
+)
+(
+x
++
+y
- )(1+y
++
+z
- +z
+)
+(
- )→
+x
-x
++
+y
++
+z
+)
+≥
+.
+x(x−1)(x
+ +x+1)⋅
+(x
   +y
   +z
+  )(x
-x
- ≤
-(x
   +y
   +z
   )
+(x−1)(y
-x(1+y
   +z
@@ Line 510: / Line 620: @@
   )
+  =
+(x
-⇒
-⇒ RHS
+ +y
-≤
-x
+ +z
-(
+ )(x
-+
-y
+ +y
+ +z
+ )
+x(x−1)
+ (x
+ +x+1)(y
+ +z
-+
+ )
-z
+ ≥0.
+Hence,
+x
+−
+x
-)
+x
 +
 y
-(
 +
 z
-+
+≥
 x
-)
+−
-+
-z
-(
-+
 x
-+
-y
-)
-(
 x
 +
 y
@@ Line 556: / Line 673: @@
 z
-)
+.
+x
+ +y
+ +z
+x
+ −x
-=
-x
-+
+ ≥
-y
+x
-+
-z
+ +y
-+
-(
+ +z
-x
-y
-+
+x
+ −x
+ .
+Similarly, we have the analogous inequalities for
 y
+y and
 z
-+
+z. Therefore, it suffices to prove that
-z
 x
-)
-(
-x
-+
-y
-+
-z
-)
 −
 x
-y
-z
-(
 x
 +
 y
@@ Line 596: / Line 718: @@
 z
-)
++
+y
+−
+y
+y
++
+z
++
+x
++
+z
+−
+z
+z
++
+x
-≤
++
-(x
+y
+≥
+,
+x
   +y
   +z
- )
-x(1+y
+x
+ −x
+ +
+y
   +z
- )+y(1+z
   +x
-  )+z(1+x
+y
- +y
+  −y
-  )
-  =
+  +
-(x
+z
+ +x
   +y
- +z
- )
-x+y+z+(xy+yz+zx)(x+y+z)−3xyz
+z
+ −z
+  ≥0,
+or equivalently,
-≤
+x
-t
+x
 +
-t
+y
++
-t
+z
++
+y
+y
-=
-t
 +
+z
-t
-−
++
+x
-t
++
+z
+z
-≤
++
+x
-⇔
-(
++
-t
+y
-−
-)
-(
-t
-−
-)
 ≥
+x
-≤
-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
@@ Line 711: / Line 828: @@
 +
-y
-−
 y
 y
@@ Line 724: / Line 838: @@
 +
-z
-−
 z
 z
@@ Line 736: / Line 847: @@
 y
-≥
+.
-.
+(1)
 x
@@ Line 747: / Line 858: @@
 x
-  −x
   +
@@ Line 759: / Line 870: @@
 y
-  −y
   +
@@ Line 771: / Line 882: @@
 z
-  −z
-  ≥0.
+  ≥
-Or equivalently,
+x
-x
+ +y
+ +z
 x
+ +
+y
-+
+ +z
-y
-+
+ +x
-z
-+
 y
+ +
+z
-y
+ +x
-+
-z
-+
+ +y
-x
-+
 z
-z
+ .(1)
+Let
-+
+t
+=
 x
 +
 y
++
-≥
+z
+t=x+y+z. Since
 x
+,
+y
+,
+z
+>
+x,y,z>0 and
 x
-+
 y
-+
 z
+≥
-+
+xyz≥1, by AM–GM we have
+t
+≥
+x
 y
+z
+≥
+t≥3
+xyz
+ ≥3. Also, note that
+x
 y
 +
+y
 z
 +
+z
 x
+≤
+t
+xy+yz+zx≤
+t
-+
-z
-z
+  and
-+
 x
@@ Line 838: / Line 977: @@
 y
-.
++
+z
+≥
+t
 x
   +y
   +z
+  ≥
-x
+t
-  +
+  .
-y
+Now, we estimate the right-hand side of (1). By the Cauchy–Schwarz inequality, we have:
- +z
+(
+x
- +x
++
+y
++
-y
+z
+)
- +
+≤
-z
+(
+x
- +x
++
+y
++
+z
+)
+(
++
+y
++
+z
+)
+,
+(x
   +y
+  +z
-z
+  )
+  ≤(x
-  ≥
-x
   +y
@@ Line 881: / Line 1,046: @@
   +z
+  )(1+y
-x
- +
-y
   +z
-  +x
+  ),
+since
-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:
 (
@@ Line 920: / Line 1,064: @@
 )
-≤
+=
 (
 x
+⋅
 +
 y
+⋅
+y
 +
 z
+⋅
+z
+)
-)
+≤
 (
+x
 +
 y
@@ Line 940: / Line 1,091: @@
 )
-⇒
+(
-x
-x
 +
 y
@@ Line 950: / Line 1,100: @@
 z
-≤
+)
+=
+(
 x
-(
 +
 y
@@ Line 962: / Line 1,113: @@
 )
 (
-x
 +
 y
@@ Line 971: / Line 1,121: @@
 )
 .
 (x
@@ Line 980: / Line 1,129: @@
   )
+ =(x
+ ⋅1+y⋅y+z⋅z)
   ≤(x
@@ Line 987: / Line 1,140: @@
   +z
-  )(1+y
+  )(1
+ +y
   +z
-  )⇒
+  )=(x
-x
   +y
@@ Line 998: / Line 1,152: @@
   +z
+  )(1+y
-x
- ≤
-(x
-  +y
+  +z
+ ).
+It follows that
+x
++
+y
++
+z
- +z
+≤
++
+y
- )
++
+z
+(
-x(1+y
+x
- +z
++
+y
- )
- .
-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
+z
-t
-/
-=
-t
-(
-t
-+
-t
 )
-=
+,
-t
+x
-⋅
+ +y
-t
+ +z
-+
-t
-=
+ ≤
+(x
+ +y
+ +z
+ )
++y
+ +z
+ ,
+and multiplying by
+x
+x (which is positive) gives:
+x
+x
++
+y
++
+z
+≤
+x
 (
-t
 +
+y
-t
-)
++
-t
+z
-=
+)
+(
-t
+x
++
+y
 +
+z
-t
-t
+)
 .
-RHS≤
+x
-t
-  /9
+  +y
-t+
+ +z
-t
+x
+ ≤
+(x
+ +y
+ +z
+ )
+x(1+y
+ +z
+ )
-  =
+  .
-t
+Similarly,
+y
+y
++
+z
- (t+
++
+x
-t
+≤
+y
+(
- )=
-t
++
+z
++
+x
+)
+(
+x
++
+y
++
+z
+)
+,
+z
+z
++
+x
- ⋅
++
+y
-t+t
- =
-t
-(t
- +3t)
- =
-t
-t
- +9t
- .
-Wait, the original says:
 ≤
-t
+z
+(
 +
-t
+x
-t
-=
-t
 +
+y
-t
-−
-t
-≤
-⇔
-(
-t
-−
 )
 (
-t
+x
-−
++
+y
++
+z
 )
-≥
+.
-≤
+y
-t
+ +z
+ +x
-t+
+y
-t
+ ≤
+(x
+ +y
+ +z
+ )
+y(1+z
+ +x
+ )
-  =
+  ,
-t
+z
+ +x
+ +y
-t
+z
- +9t−27
-  ≤1⇔(t−3)(t
+  ≤
+(x
-  −9)≥0
-But there is a discrepancy: The step from the previous expression to the next is not clear. Actually, the original says:
+  +y
-≤
+ +z
-t
-+
+  )
-t
-t
-=
-t
-+
-t
-−
-t
-≤
-t
-t+
-t
+z(1+x
+ +y
+ )
+  .
+Summing these, we obtain:
- =
-t
+∑
+x
+x
++
-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
@@ Line 1,275: / Line 1,432: @@
 )
+]
-≤
+.
-t
+∑
+x
+ +y
+ +z
+x
+ ≤
+(x
+ +y
+ +z
+ )
+ [∑x+∑x(y
+ +z
+ )].
+Now, note that
+∑
+x
+(
+y
 +
-(
+z
-t
-/
 )
-t
-t
-/
 =
-t
+∑
+(
+x
+y
 +
-t
+x
+z
-/
+)
-t
-/
 =
+(
-t
+x
+y
 +
-t
+x
+z
-/
+)
-t
-=
-t
 +
+(
-t
+y
+z
-t
-=
-t
 +
+y
-t
+x
-t
-.
-(x
- +y
+)
++
+(
+z
+x
- +z
++
+z
+y
-  )
+)
+=
+(
+x
++
+y
++
+z
+)
+(
+x
+y
++
+y
+z
++
+z
+x
+)
+−
+x
+y
+z
+.
+∑x(y
+ +z
+  )=∑(xy
+ +xz
+  )=(xy
-x+y+z+(xy+yz+zx)(x+y+z)−3xyz
- ≤
-t
- /9
-t+(t
-  /3)t
+  +xz
-  =
+ )+(yz
-t
+ +yx
+ )+(zx
+  +zy
+ )=(x+y+z)(xy+yz+zx)−3xyz.
+Thus,
+∑
+x
+x
- /9
++
-t+t
+y
++
+z
+≤
+t
++
+t
+(
+x
+y
++
+y
+z
++
+z
+x
+)
+−
- /3
+x
+y
- =9
+z
-t
+(
+x
++
+y
++
+z
+)
+.
+∑
+x
+ +y
+ +z
-t+t
+x
- /3
-  =
+  ≤
-t
+(x
+  +y
-t+3t
+ +z
+ )
+t+t(xy+yz+zx)−3xyz
-  =
+  .
-t
+Since
+x
+y
-t
+z
+≥
+xyz≥1, we have
+−
- +9t
+x
+y
- .
+z
-But then the original writes:
 ≤
-t
+−
+−3xyz≤−3. Also,
+x
+y
++
+y
+z
 +
+z
+x
+≤
 t
+xy+yz+zx≤
+t
+  and
+x
++
+y
++
+z
+≥
 t
-=
-t
-+
+x
-t
+ +y
-−
+ +z
-t
+  ≥
-≤
-t
-t+
 t
+  , so
- =
+(
-t
+x
++
-t
+y
- +9t−27
-That is inconsistent. Possibly there is a misprint. Actually, the original then says:
-t
 +
+z
-t
-−
+)
-t
+≥
-≤
-⇔
 (
 t
-−
 )
-(
+=
 t
-−
+.
+(x
+ +y
+ +z
+ )
+ ≥(
+t
+ )
+ =
-)
-≥
-.
 t
-t
- +9t−27
-  ≤1⇔(t−3)(t
+  .
+Therefore,
- −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}
+x
-\leq \frac{t + \frac{t^3}{3}}{t^4} = \frac{3t^3 + 9t - 27}{t^4} \leq 1</math>"
+x
-Wait, the original has:
-≤
-t
 +
-t
+y
++
+z
+≤
+t
++
+t
+⋅
+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
@@ Line 1,522: / Line 1,782: @@
 −
-(
-x
-+
-y
-+
-z
 )
+t
-≤
+=
 t
 +
 t
 −
-t
-/
-=
-(
-t
-+
-t
-/
-−
-)
-t
-=
-t
-+
-t
-−
 t
@@ Line 1,581: / Line 1,807: @@
 t
-≤
+.
-(x
+∑
+x
   +y
   +z
- )
-t+
+x
+ ≤
+(t
+ /3)
+t+t⋅
 t
   −3
-  ≤
+  =
 t
@@ Line 1,615: / Line 1,848: @@
-(t+t
+(t+
-  /3−3)
+t
+ −3)
   =
@@ Line 1,635: / Line 1,872: @@
   +9t−27
+  .
-because we also subtract 3xyz? Actually, note that the numerator is:
+We now show that
-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
@@ Line 1,657: / Line 1,885: @@
 t
-.
+≤
-RHS≤
+.
 t
@@ Line 1,666: / Line 1,894: @@
   +9t−27
-  .
+  ≤1.
-Then we want to show that this is ≤ 1, i.e.,
+This is equivalent to
-t
-+
-t
-−
-t
-≤
-⇔
 t
@@ Line 1,707: / Line 1,922: @@
 ≥
 .
 t
-t
- +9t−27
- ≤1⇔3t
   +9t−27≤t
@@ Line 1,723: / Line 1,931: @@
   −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.
+Factor the left-hand side:
-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.
+t
-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
+t
-x
+−
+t
 +
-y
+=
-+
+(
-z
+t
+−
-≥
-x
+)
+(
+t
 −
-x
-x
+)
+.
+t
-+
+ −3t
-y
+ −9t+27=(t−3)(t
+ −9).
+For
+t
+≥
+t≥3, we have
+t
+−
+≥
+t−3≥0 and
+t
+−
+≥
+t
+ −9≥18, so indeed
+(
+t
+−
+)
+(
+t
+−
+)
+≥
+(t−3)(t
+ −9)≥0. Hence,
+∑
+x
+x
++
+y
 +
 z
-x
+≤
+.
- +y
+(2)
+∑
- +z
-x
- −x
- ≥
 x
@@ Line 1,895: / Line 2,022: @@
 x
- −x
+  ≤1.(2)
-since
+Now, we estimate the left-hand side of (1). Note that
 x
-−
 x
-x
 +
 y
@@ Line 1,915: / Line 2,037: @@
 z
-−
+=
 x
-−
 x
++
 x
-+
 y
 +
-z
-=
-x
-(
 x
-−
-)
-(
-x
-+
-x
-+
-)
-(
-y
-+
 z
-)
+,
-(
+x
-x
+ +y
-+
-y
-+
+ +z
-z
-)
-(
+x
-x
-+
-y
+ =
+x
+ +x
-+
+ y
-z
-)
+  +x
-≥
-,
-x
-  +y
-  +z
+  z
 x
- −x
-  −
+  ,
-x
+and similarly for the other terms. So,
+∑
+x
+x
- +y
++
+y
++
+z
- +z
+=
+∑
+x
+x
++
+x
+y
++
+x
+z
+.
+∑
 x
- −x
- =
-(x
   +y
   +z
-  )(x
+x
-  +y
-  +z
+  =∑
+x
-  )
-x(x−1)
+  +x
-  (x
+  y
-  +x+1)(y
+  +x
-  +z
+  z
-  )
+x
-  ≥0,
+  .
-which holds for all positive real numbers <math>x, y, z</math>.
+By the Cauchy–Schwarz inequality (Titu's lemma), we have:
-Thus, it is enough to prove that
+∑
 x
-−
 x
++
 x
-+
 y
 +
+x
 z
+≥
+(
+x
 +
 y
-−
-y
-y
 +
 z
+)
+∑
+(
+x
 +
 x
+y
 +
+x
 z
-−
+)
+=
+(
+x
++
+y
++
 z
-z
+)
-+
 x
 +
 y
++
+z
++
+(
+x
+y
++
+y
+z
++
+z
-≥
+x
-.
-x
- +y
- +z
+)
+.
+(3)
+∑
 x
-  −x
+  +x
-  +
+  y
-y
- +z
   +x
+ z
+x
-y
- −y
-  +
+ ≥
-z
+∑(x
+  +x
+ y
   +x
+ z
+ )
+(x
   +y
+ +z
+ )
-z
- −z
-  ≥0.
+  =
-Rewriting, we need to show
+x
-x
+ +y
-x
+ +z
-+
+ +2(x
-y
-+
+ y
-z
-+
+ +y
-y
-y
-+
-z
-+
+ z
-x
-+
+ +z
-z
-z
-+
-x
-+
+ x
-y
-≥
+ )
+(x
+ +y
+ +z
+ )
+ .(3)
+We claim that
+(
 x
-x
 +
 y
 +
 z
+)
+≥
+x
 +
 y
-y
 +
 z
++
-+
+(
 x
-+
+y
-z
-z
-+
-x
 +
 y
+z
++
+z
+x
+)
 .
-x
+(x
   +y
+ +z
+ )
+ ≥x
+ +y
   +z
+ +2(x
+ y
+ +y
+ z
-x
- +
-y
   +z
-  +x
+  x
+  ).
-y
+Expanding the left-hand side:
+(
+x
- +
-z
++
+y
- +x
++
+z
+)
- +y
+=
+x
++
+y
++
+z
++
+(
-z
+x
+y
- ≥
++
-x
+y
+z
++
+z
+x
+)
+.
+(x
   +y
   +z
+ )
+  =x
-x
+  +y
-  +
-y
   +z
-  +x
+  +2(x
+  y
-y
- +
-z
- +x
   +y
+  z
-z
+ +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>.
+ x
+  ).
+Thus, the inequality is equivalent to
-Now, estimate the right-hand side (RHS). By the Cauchy–Schwarz inequality,
-(
 x
+y
 +
 y
+z
 +
 z
+x
+≥
+x
-)
+y
-≤
-(
-x
 +
 y
-+
 z
-)
-(
-+
-y
 +
 z
-)
+x
-,
-(x
+.
+(4)
+x
+ y
   +y
+ z
   +z
+ x
+ ≥x
-  )
+  y
- ≤(x
   +y
+ z
   +z
-  )(1+y
+  x
-  +z
+  .(4)
+Let
- ),
+a
-so
+=
+x
+y
+a=xy,
+b
+=
+y
+z
+b=yz,
+c
+=
+z
 x
+c=zx. Then
+a
+b
+c
+=
 x
-+
 y
-+
 z
-≤
+≥
-x
-(
+abc=x
+ y
+ z
+ ≥1, and inequality (4) becomes:
+a
 +
-y
+b
 +
-z
+c
-)
+≥
-(
+a
-x
 +
-y
+b
 +
-z
+c
-)
 .
-x
+a
-  +y
+  +b
+ +c
+ ≥a
-  +z
+  +b
+  +c
-x
- ≤
-(x
- +y
- +z
- )
-x(1+y
- +z
- )
   .
-Cyclically summing, we get
+We now prove this. By the AM–GM inequality,
-\begin{align*}
+a
-\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
++
+b
 +
+c
-t
-−
+≥
-t
+a
+b
-≤
+c
-.
+≥
-t
+a
-t
+  +b
+ +c
-  +9t−27
+  ≥3abc≥3. Also, by the power mean inequality, we have:
- ≤1.
-This is equivalent to
+(
+a
-t
++
+b
 +
+c
-t
-−
+)
-≤
-t
+/
-⇔
-t
-−
-t
+≥
+(
-−
+a
-t
++
+b
 +
+c
-≥
-.
-t
- +9t−27≤t
+)
- ⇔t
+/
- −3t
+,
+(
- −9t+27≥0.
+a
-Factoring, we have
-t
-−
-t
+ +b
-−
+ +c
-t
-+
-=
-(
-t
-−
-)
-(
-t
+ )
+/3
+ ≥(
-−
+a
-)
+ +b
-≥
+ +c
-,
-t
- −3t
+ )
+/2
+ ,
+which implies
+a
- −9t+27=(t−3)(t
++
+b
- −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
+c
+≥
+(
+a
 +
-z
+b
-=
-x
-x
 +
-x
+c
-y
+)
+/
+=
+(
+a
++
+b
 +
-x
+c
-z
+)
+/
 .
-x
+a
-  +y
+  +b
+ +c
+ ≥3(
+a
+ +b
-  +z
+  +c
-x
+ )
+/2
   =
-x
-  +x
+(a
-  y
+  +b
-  +x
+  +c
- z
+  )
-x
+/2
   .
-Thus,
+Squaring both sides (all positive) gives:
-LHS
+(
-=
+a
-∑
-x
-x
 +
-y
+b
 +
-z
+c
+)
+≥
+(
+a
-=
-∑
-x
-x
 +
-x
+b
-y
 +
-x
+c
-z
+)
+.
+(a
+ +b
+ +c
+ )
-.
+ ≥
-LHS=∑
-x
+(a
- +y
-  +z
+  +b
+  +c
-x
- =∑
-x
-  +x
- y
- +x
- z
+  )
-x
   .
-By the Cauchy–Schwarz inequality,
+That is,
 (
-∑
+a
-x
-x
 +
-x
+b
-y
 +
-x
+c
-z
 )
+≤
 (
-∑
+a
-(
-x
 +
-x
+b
-y
 +
-x
+c
-z
-)
-)
-≥
-(
-∑
-x
 )
 .
-(∑
+(5)
-x
+(a
- +x
-  y
+  +b
-  +x
+  +c
-  z
+  )
+  ≤3(a
-x
+  +b
+  +c
- )(∑(x
-  +x
- y
-  +x
- z
- ))≥(∑x
   )
-  .
+  .(5)
-But
+On the other hand, since
+a
-∑
-(
-x
 +
-x
+b
-y
 +
-x
+c
-z
+≥
-)
+a
-=
-∑
+ +b
-x
+ +c
+ ≥3, we have
+(
+a
 +
-∑
+b
-x
-y
 +
-∑
+c
-x
+)
-z
+≤
+(
-=
+a
-∑
-x
++
+b
++
+c
+)
+,
+because
+(
+a
++
+b
++
+c
+)
+≥
+(
+a
++
+b
 +
+c
+)
-∑
+⇔
-x
+a
-y
-.
-∑(x
- +x
- y
- +x
- z
- )=∑x
- +∑x
- y
- +∑x
- z
- =∑x
- +2∑x
- y
- .
-Hence,
-LHS
-≥
-(
-x
 +
-y
+b
 +
-z
+c
+≥
+.
+(a
+ +b
+ +c
-)
+ )
-x
+ ≤(a
-+
+ +b
-y
+ +c
-+
-z
+ )
-+
+ ,because(a
-(
+ +b
-x
+ +c
-y
+ )
+ ≥3(a
+ +b
+ +c
+ )
+ ⇔a
+ +b
+ +c
+ ≥3.
+Combining with (5), we get:
+(
+a
 +
-y
+b
-z
 +
-z
+c
-x
 )
-.
-LHS≥
+≤
-x
+(
+a
- +y
++
- +z
+b
- +2(x
++
+c
+)
+,
+(a
-  y
+  +b
-  +y
+  +c
-  z
+  )
-  +z
+  ≤(a
- x
- )
-(x
-  +y
+  +b
-  +z
+  +c
   )
+ ,
+so taking cube roots yields
+a
++
+b
- .
-We now show that this expression is at least 1. That is, we need
++
+c
-(
-x
+≤
+a
 +
-y
+b
 +
-z
+c
-)
+a
+ +b
+ +c
-≥
+ ≤a
+ +b
+ +c
+ , as desired.
+Therefore, from (3) we obtain
+∑
+x
 x
 +
 y
 +
 z
-+
-(
+≥
+.
+(6)
+∑
+x
+ +y
+ +z
+x
+ ≥1.(6)
+Combining (2) and (6), we have
+∑
+x
 x
++
+y
-y
++
+z
+≥
+≥
+∑
+x
+x
 +
 y
-z
 +
 z
-x
+,
+∑
-)
+x
-.
-(x
   +y
- +z
- )
- ≥x
- +y
   +z
- +2(x
- y
+x
+ ≥1≥∑
+x
   +y
- z
   +z
-  x
+x
-  ).
-Expanding the left-hand side:
+  ,
+which proves (1). Hence, the original inequality holds.
-(
+Equality occurs when
 x
+=
-+
 y
+=
-+
 z
-)
 =
-x
+x=y=z=1.
-+
-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}