Difference between revisions of "2012 USAJMO Problems/Problem 3"

(Created page with "== Problem == Let <math>a</math>, <math>b</math>, <math>c</math> be positive real numbers. Prove that <cmath>\frac{a^3 + 3b^3}{5a + b} + \frac{b^3 + 3c^3}{5b + c} + \frac{c^3 +...")
 
(Solution)
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==Solution==
 
==Solution==
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By the Cauchy-Schwarz inequality,
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<cmath>[a(5a + b) + b(5b + c) + c(5c + a)] \left( \frac{a^3}{5a + b} + \frac{b^3}{5b + c} + \frac{c^3}{5c + a} \right) \ge (a^2 + b^2 + c^2)^2,</cmath>
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so
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<cmath>\frac{a^3}{5a + b} + \frac{b^3}{5b + c} + \frac{c^3}{5c + a} \ge \frac{(a^2 + b^2 + c^2)^2}{5a^2 + 5b^2 + 5c^2 + ab + ac + bc}.</cmath>
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Since <math>a^2 + b^2 + c^2 \ge ab + ac + bc</math>,
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<cmath>\frac{(a^2 + b^2 + c^2)^2}{5a^2 + 5b^2 + 5c^2 + ab + ac + bc} \ge \frac{(a^2 + b^2 + c^2)^2}{6a^2 + 6b^2 + 6c^2} = \frac{1}{6} (a^2 + b^2 + c^2).</cmath>
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Hence,
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<cmath>\frac{a^3}{5a + b} + \frac{b^3}{5b + c} + \frac{c^3}{5c + a} \ge \frac{1}{6} (a^2 + b^2 + c^2).</cmath>
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 +
By the same argument,
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<cmath>\frac{b^3}{5a + b} + \frac{c^3}{5b + c} + \frac{a^3}{5c + a} \ge \frac{1}{6} (a^2 + b^2 + c^2).</cmath>
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Therefore,
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<cmath>\frac{a^3 + 3b^3}{5a + b} + \frac{b^3 + 3c^3}{5b + c} + \frac{c^3 + 3a^3}{5c + a} \ge \frac{1 + 3}{6} (a^2 + b^2 + c^2) = \frac{2}{3} (a^2 + b^2 + c^2).</cmath>
  
 
==See Also==
 
==See Also==

Revision as of 18:27, 24 April 2012

Problem

Let $a$, $b$, $c$ be positive real numbers. Prove that \[\frac{a^3 + 3b^3}{5a + b} + \frac{b^3 + 3c^3}{5b + c} + \frac{c^3 + 3a^3}{5c + a} \ge \frac{2}{3} (a^2 + b^2 + c^2).\]

Solution

By the Cauchy-Schwarz inequality, \[[a(5a + b) + b(5b + c) + c(5c + a)] \left( \frac{a^3}{5a + b} + \frac{b^3}{5b + c} + \frac{c^3}{5c + a} \right) \ge (a^2 + b^2 + c^2)^2,\] so \[\frac{a^3}{5a + b} + \frac{b^3}{5b + c} + \frac{c^3}{5c + a} \ge \frac{(a^2 + b^2 + c^2)^2}{5a^2 + 5b^2 + 5c^2 + ab + ac + bc}.\] Since $a^2 + b^2 + c^2 \ge ab + ac + bc$, \[\frac{(a^2 + b^2 + c^2)^2}{5a^2 + 5b^2 + 5c^2 + ab + ac + bc} \ge \frac{(a^2 + b^2 + c^2)^2}{6a^2 + 6b^2 + 6c^2} = \frac{1}{6} (a^2 + b^2 + c^2).\] Hence, \[\frac{a^3}{5a + b} + \frac{b^3}{5b + c} + \frac{c^3}{5c + a} \ge \frac{1}{6} (a^2 + b^2 + c^2).\]

By the same argument, \[\frac{b^3}{5a + b} + \frac{c^3}{5b + c} + \frac{a^3}{5c + a} \ge \frac{1}{6} (a^2 + b^2 + c^2).\] Therefore, \[\frac{a^3 + 3b^3}{5a + b} + \frac{b^3 + 3c^3}{5b + c} + \frac{c^3 + 3a^3}{5c + a} \ge \frac{1 + 3}{6} (a^2 + b^2 + c^2) = \frac{2}{3} (a^2 + b^2 + c^2).\]

See Also

2012 USAJMO (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6
All USAJMO Problems and Solutions
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