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

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<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]].
 
<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]].
 
{{IMO box|year=2001|num-b=1|num-a=3}}
 
{{IMO box|year=2001|num-b=1|num-a=3}}
[[Category:Olympiad Inequality Problems]]
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[[Category:Olympiad Number Theory Problems]]

Revision as of 19:57, 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.

2001 IMO (Problems) • Resources
Preceded by
Problem 1
1 2 3 4 5 6 Followed by
Problem 3
All IMO Problems and Solutions