2005 IMO Problems/Problem 3

\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

) 

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

) 

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

) 

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

) 

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

) 

2

y(1+z 2

+x 

2

)

​

, 

z 4

+x 

2

+y 

2

z ​

≤ 

(x 2

+y 

2

+z 

2

) 

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

) 

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

) 

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

) 

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

y 

2

+x 

2

z 

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

y 

2

+x 

2

z 

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

y 

2

+x 

2

z 

2

x 6

​

≥ 

∑(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

)

(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

+y 

2

z 

2

+z 

2

x 

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

+y 

3

z 

3

+z 

3

x 

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

y 

3

+y 

3

z 

3

+z 

3

x 

3

≥x 

2

y 

2

+y 

2

z 

2

+z 

2

x 

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

y 

2

z 

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

) 

3

,because(a 

3

+b 

3

+c 

3

) 

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

) 

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}