1997 USAMO Problems/Problem 5: Difference between revisions

Latest revision as of 21:06, 28 December 2024

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 $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}$

Solution 3

If we multiply each side by $abc$ , we get that we must just prove that $\frac{abc}{a^3+b^3+abc} + \frac{abc}{b^3+c^3+abc} + \frac{abc}{c^3+a^3+abc} \leq 1$ If we divide our LHS equation, we get that $\frac{1}{\frac{a^3+b^3+abc}{abc}} +\frac{1}{\frac{b^3+c^3+abc}{abc}} + \frac{1}{\frac{c^3+a^3+abc}{abc}}$ $= \frac{1}{\frac{a^2}{bc} + \frac{b^2}{ac} + 1} +\frac{1}{\frac{b^2}{ac} + \frac{c^2}{ab} + 1} + \frac{1}{\frac{c^2}{ab} + \frac{a^2}{bc} + 1}$ Make the astute observation that by Titu's Lemma, $\frac{a^2}{bc} + \frac{b^2}{ac} \geq \frac{\left(a+b\right)^2}{bc+ ac}$ $\sum_{cyc}\frac{\left(a+b\right)^2}{bc+ ac} \leq \sum_{cyc}\frac{a^2}{bc} + \frac{b^2}{ac}$ Therefore: $\sum_{cyc} \frac{1}{\frac{\left(a+b\right)^2}{c\left(a+b\right) }+1} \geq \sum_{cyc} \frac{1}{\frac{a^2}{bc} + \frac{b^2}{ac} + 1}$ If we expand it out, we get that $\sum_{cyc} \frac{1}{\frac{\left(a+b\right)^2}{c\left(a+b\right) }+1} = \frac{a+b+c}{a+b+c} = 1$ Since our original equation is less than this, we get that $\sum_{cyc} \frac{1}{\frac{a^2}{bc} + \frac{b^2}{ac} + 1} \leq 1$ $\sum_{cyc} \frac{abc}{a^3+b^3+abc}\leq 1$ $\sum_{cyc} \frac{1}{a^3+b^3+abc} \leq \frac{1}{abc}$

-KEVIN_LIU

Video Solution (inspired by Solution 1)

https://youtu.be/6czJm7FMGtk

@@ Line 2: / Line 2: @@
 Prove that, for all positive real numbers <math>a, b, c,</math>
-<math>\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}</math>.
+<cmath>\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}</cmath>.
 == Solution 1 ==
@@ Line 28: / Line 28: @@
 -KEVIN_LIU
+== Video Solution (inspired by Solution 1) ==
+https://youtu.be/6czJm7FMGtk
 ==See Also ==

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

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