Difference between revisions of "2001 IMO Problems/Problem 2"

Revision as of 17:45, 25 October 2007

Problem

Let $a,b,c$ be positive real numbers. Prove that $\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1$

Solution

Solution using Holder's

By Holder's inequality, $\left(\sum\frac{a}{\sqrt{a^{2}+8bc}}\right)\left(\sum\frac{a}{\sqrt{a^{2}+8bc}}\right)\left(\sum a(a^{2}+8bc)\right)\ge (a+b+c)^{3}$ Thus we need only show that $(a+b+c)^{3}\ge a^{3}+b^{3}+c^{3}+24abc$ Which is obviously true since $(a+b)(b+c)(c+a)\ge 8abc$ .

Alternate Solution using Jensen's

By Jensen's, $\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge a^{3}+b^{3}+c^{3}+24abc$ , so we need to prove $a^{3}+b^{3}+c^{3}+24abc\ge 1$ , which is obvious by RMS-AM-GM-HM.

@@ Line 11: / Line 11: @@
 ===Alternate Solution using Jensen's===
 By Jensen's,
-<math>\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge a^{3}+b^{3}+c^{3}+24abc</math>, so we need to prove <math>a^{3}+b^{3}+c^{3}+24abc\ge 1</math>, which is obvious by [[AMGM]].
+<math>\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge a^{3}+b^{3}+c^{3}+24abc</math>, so we need to prove <math>a^{3}+b^{3}+c^{3}+24abc\ge 1</math>, which is obvious by [[RMS-AM-GM-HM]].

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