Difference between revisions of "2001 IMO Problems/Problem 2"
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| − | ==Problem== | + | == Problem == |
| − | Let <math>a,b,c</math> be positive real numbers. Prove that | + | Let <math>a,b,c</math> be positive real numbers. Prove that <math>\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1</math>. |
| − | <math>\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1</math> | + | |
| − | ==Solution== | + | __TOC__ |
| − | ===Solution using Holder's=== | + | == Solution == |
| + | === Solution using Holder's === | ||
By Holder's inequality, | By Holder's inequality, | ||
<math>\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}</math> | <math>\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}</math> | ||
Thus we need only show that | Thus we need only show that | ||
<math>(a+b+c)^{3}\ge a^{3}+b^{3}+c^{3}+24abc</math> | <math>(a+b+c)^{3}\ge a^{3}+b^{3}+c^{3}+24abc</math> | ||
| − | Which is obviously true since <math>(a+b)(b+c)(c+a)\ge 8abc</math>. | + | Which is obviously true since <math>(a+b)(b+c)(c+a)\ge 8abc</math>. |
| − | ===Alternate Solution using Jensen's=== | + | |
| + | === Alternate Solution using 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: | 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: | ||
<cmath>\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}}</cmath> | <cmath>\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}}</cmath> | ||
but | but | ||
<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. | <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. | ||
| + | |||
| + | == See also == | ||
| + | {{IMO box|year=2001|num-b=1|num-a=3}} | ||
| + | |||
| + | [[Category:Olympiad Inequalitiy Problems]] | ||
Revision as of 21:57, 21 April 2008
Problem
Let 
be positive real numbers. Prove that 
.
Contents
Solution
Solution using Holder's
By Holder's inequality,
Thus we need only show that
Which is obviously true since 
.
Alternate Solution using Jensen's
This inequality is homogeneous so we can assume without loss of generality 
and apply Jensen's inequality for 
, so we get:
![\[\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ac}}+\frac{c}{\sqrt{c^2+8ab}} \geq \frac{1}{\sqrt{a^3+b^3+c^3+24abc}}\]](http://latex.artofproblemsolving.com/3/9/e/39e0afe795d1a4d834a9d2489badc1fcfb9fcb69.png)
but
![\[1=(a+b+c)^3=a^3+b^3+c^3+6abc+3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) \geq a^3+b^3+c^3+24abc\]](http://latex.artofproblemsolving.com/7/9/5/79522da167994a2759a51bc68bf0790cce7d2af8.png)
, and thus the inequality is proven.
See also
| 2001 IMO (Problems) • Resources | ||
| Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
| All IMO Problems and Solutions | ||
