AoPSWiki
Our Precalculus course starts on Dec. 4. Master trig, complex numbers, and vectors and matrices in 2 and 3 dimensions. Click here to enroll today!

2001 IMO Problems/Problem 2

From AoPSWiki

Revision as of 18:11, 27 June 2008 by Temperal (Talk | contribs)
(diff) ← Older revision | Current revision (diff) | Newer revision → (diff)

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.

Contents

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+... 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

This inequality is homogeneous so we can assume without loss of generality a+b+c=1 and apply Jensen's inequality for f(x)=\frac{1}{\sqrt{x}}, so we get: \frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ac}}+\frac{c}{\sqrt{c^2+8ab}} \geq \frac{1}{\sqrt{a^3+b^3+c^3+24abc}} but 1=(a+b+c)^3=a^3+b^3+c^3+6abc+3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) \geq a^3+b^3+c^3+24abc by AMGM, and thus the inequality is proven.

See also

2001 IMO (Problems)
Preceded by
Problem 1
1 2 3 4 5 6 Followed by
Problem 3
Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us