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

Revision as of 19:44, 23 November 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

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: $\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ac}}+\frac{b}{\sqrt{b^2+8ac}} \geq \frac{1}{\sqrt{a^3+b^3+c^3+24abc}}$ but $1=(a+b+c)^3=a^3+b^3+c^3+6abc+3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) \geq a^3+b^3+c^3+24abc$ , and thus the inequality is proven.

@@ Line 10: / Line 10: @@
 Which is obviously true since <math>(a+b)(b+c)(c+a)\ge 8abc</math>.
 ===Alternate Solution using Jensen's===
-By Jensen's,
+This inequality is homogeneous so we can assume without loss of generality <math>a+b+c=1</math> and apply Jensen's inequality for <math>f(x)=\frac{1}{\sqrt{x}}</math>, so we get:
-<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>\frac{1}{\sqrt{a^{3}+b^{3}+c^{3}+24abc}}\ge 1</math>, which is obvious by [[RMS-AM-GM-HM]].
+<cmath>\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ac}}+\frac{b}{\sqrt{b^2+8ac}} \geq \frac{1}{\sqrt{a^3+b^3+c^3+24abc}}</cmath>
-{{IMO box|year=2001|num-b=1|num-a=3}}
+but
-[[Category:Olympiad Number Theory Problems]]
+<cmath>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</cmath>, and thus the inequality is proven.

Page

Toolbox

Search