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}
