2004 USAMO Problems/Problem 5

Problem 5

(Titu Andreescu) Let $a$ , $b$ , and $c$ be positive real numbers. Prove that

$(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \ge (a+b+c)^3$ .

Solutions

We first note that for positive $x$ , $x^5 + 1 \ge x^3 + x^2$ . We may prove this in the following ways:

Since $x^2$ and $x^3$ must be both lesser than, both equal to, or both greater than 1, by the rearrangement inequality, $x^2 \cdot x^3 + 1 \cdot 1 \ge x^2 \cdot 1 + 1 \cdot x^3$ .

Since $x^2 - 1$ and $x^3 - 1$ have the same sign, $0 \le (x^2 - 1)(x^3 - 1) = x^5 - x^3 - x^2 + 1$ , with equality when $x = 1$ .

By weighted AM-GM, $\frac{2}{5}x^5 + \frac{3}{5} \ge x^2$ and $\frac{3}{5}x^5 + \frac{2}{5} \ge x^3$ . Adding these gives the desired inequality. Equivalently, the desired inequality is a case of Muirhead's Inequality.

It thus becomes sufficient to prove that

$(a^3 + 2)(b^3 + 2)(c^3 + 2) \ge (a+b+c)^3$ .

We present two proofs of this inequality:

By Hölder's Inequality,

$\begin{matrix}(m_{1,1} + m_{1,2} + m_{1,3})(m_{2,1} + m_{2,2} + m_{2,3})(m_{3,1} + m_{3,2} + m_{3,3}) \ge \\ \qquad \qquad \qquad\left[ (m_{1,1}m_{2,1}m_{3,1})^{1/3} + (m_{2,1}m_{2,2}m_{2,3})^{1/3} + (m_{3,1}m_{3,2}m_{3,3})^{1/3} \right] ^3.\end{matrix}$

We get the desired inequality by taking $m_{1,1} = a^3$ , $m_{2,2} = b^3$ , $m_{3,3} = c^3$ , and $m_{x,y} = 1$ when $x \neq y$ . We have equality if and only if $a = b = c = 1$ .

Take $x = \sqrt{a}$ , $y = \sqrt{b}$ , and $z = \sqrt{c}$ . Then some two of $x$ , $y$ , and $z$ are both at least $1$ or both at most $1$ . Without loss of generality, say these are $x$ and $y$ . Then the sequences $(x, 1, 1)$ and $(1, 1, y)$ are oppositely sorted, yielding

$(x^6 + 1 + 1)(1 + 1 + y^6) \ge 3(x^6 + 1 + y^6)$

by Chebyshev's Inequality. By the Cauchy-Schwarz Inequality we have

$(x^6 + 1 + y^6)(1 + z^6 + 1) \ge (x^3 + y^3 + z^3)^2.$

Applying Chebyshev's and the Cauchy-Schwarz Inequalities each once more, we get

$3(x^3 + y^3 + z^3) \ge (x^2 + y^2 + z^2)(x+y+z),$

and

$(x^3 + y^3 + z^3)(x+y+z) \ge (x^2 + y^2 + z^2)^2.$

Multiplying the above four inequalities together yields

$(x^6 + 2)(y^6 + 2)(z^6 + 2) \ge (x^2 + y^2 + z^2)^3,$

as desired, with equality if and only if $x = y = z = 1$ .

First, expand the left side of the inequality to get

$a^3b^3c^3 + 4(a^3 + b^3 + c^3) + 2(a^3b^3 + a^3c^3 + b^3c^3) + 8.$

By the AM-GM inequality, it is true that $a^3 + a^3b^3 + 1 \ge 3a^2b$ , and so it is clear that

$a^3b^3c^3 + 4(a^3 + b^3 + c^3) + 2(a^3b^3 + a^3c^3 + b^3c^3) + 8 \ge a^3b^3c^3 + 2(a^3 + b^3 + c^3) + 2 + 3a^2b + 3ab^2 + 3bc^2 + 3b^2c + 3c^2a + 3ca^2.$

Additionally, again by AM-GM, it is true that $a^3b^3c^3 + a^3 + b^3 + c^3 + 1 + 1 \ge 6abc$ , and so

$a^3b^3c^3 + 2(a^3 + b^3 + c^3) + 2 + 3a^2b + 3ab^2 + 3bc^2 + 3b^2c + 3c^2a + 3ca^2$

$\ge a^3 + b^3 + c^3 + 3a^2b + 3ab^2 + 3bc^2 + 3b^2c + 3c^2a + 3ca^2 + 6abc = {(a + b + c)}^3,$

as desired.

It is also possible to solve this inequality by expanding terms and applying brute force, either before or after proving that $x^5 - x^2 + 3 \ge x^3 + 2$ .

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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2004 USAMO (Problems • Resources)
Preceded by Problem 4	Followed by Problem 6
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2004 USAMO Problems/Problem 5

Problem 5

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