1997 USAMO Problems/Problem 5

Problem

Prove that, for all positive real numbers $a, b, c,$

$\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{a^3+c^3+abc} \le \frac{1}{abc}$ .

Solution 1

Because the inequality is homogenous (i.e. $(a, b, c)$ can be replaced with $(ka, kb, kc)$ without changing the inequality other than by a factor of $k^n$ for some $n$ ), without loss of generality, let $abc = 1$ .

Lemma: $\frac{1}{a^3 + b^3 + 1} \le \frac{c}{a + b + c}.$ Proof: Rearranging gives $(a^3 + b^3) c + c \ge a + b + c$ , which is a simple consequence of $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ and $(a^2 - ab + b^2)c \ge (2ab - ab)c = abc = 1.$

Thus, by $abc = 1$ : $\frac{1}{a^3 + b^3 + abc} + \frac{1}{b^3 + c^3 + abc} + \frac{1}{c^3 + a^3 + abc}$ $\le \frac{c}{a + b + c} + \frac{a}{a + b + c} + \frac{b}{a + b + c} = 1 = \frac{1}{abc}.$

Solution 2

Rearranging the AM-HM inequality, we get $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \le \frac{9}{x+y+z}$ . Letting $x=a^{3}+b^{3}+abc$ , $y=b^{3}+c^{3}+abc$ , and $z=c^{3}+a^{3}+abc$ , we get $\frac{1}{a^{3}+b^{3}+abc}+\frac{1}{b^{3}+c^{3}+abc}+\frac{1}{c^{3}+a^{3}+abc} \le \frac{9}{2a^{3}+2b^{3}+2c^{3}+3abc}.$ By AM-GM on $a^{3}$ , $b^{3}$ , and $c^{3}$ , we have $a^{3}+b^{3}+c^{3} \ge 3abc \Rightarrow 2a^{3}+2b^{3}+2c^{3}+3abc \ge 9abc \Rightarrow \frac{9}{2a^{3}+2b^{3}+2c^{3}+3abc} \le \frac{1}{abc}.$ So, $\frac{1}{a^{3}+b^{3}+abc}+\frac{1}{b^{3}+c^{3}+abc}+\frac{1}{c^{3}+a^{3}+abc} \le \frac{1}{abc}$ . -Tigerzhang

$\textbf{WARNING:}$

This solution doesn’t work because $(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\geq 9$ , so $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}$

1997 USAMO (Problems • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6
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1997 USAMO Problems/Problem 5

Contents

Problem

Solution 1

Solution 2

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