Difference between revisions of "2001 IMO Problems/Problem 2"
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<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> by AMGM, 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> by AMGM, and thus the inequality is proven. | ||
+ | |||
+ | === Alternate Solution using Isolated Fudging === | ||
+ | We claim that | ||
+ | <cmath>\frac{a}{\sqrt{a^2+8bc}} \geq \frac{a^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}</cmath> | ||
+ | Cross-multiplying, squaring both sides and expanding, we have | ||
+ | <cmath>a^{\frac{14}{3}}+a^{2}b^{\frac{8}{3}}+a^{2}c^{\frac{8}{3}}+2a^{\frac{11}{3}}b^{\frac{4}{3}}+2a^{\frac{11}{3}}c^{\frac{4}{3}}+2a^{2}b^{\frac{4}{3}}c^{\frac{4}{3}} \geq a^{\frac{14}{3}}+8a^{\frac{8}{3}}bc</cmath> | ||
+ | After cancelling the <math> a^{\frac{14}{3}}</math> term, we apply AM-GM to RHS and obtain | ||
+ | <cmath>a^{2}b^{\frac{8}{3}}+a^{2}c^{\frac{8}{3}}+2a^{\frac{11}{3}}b^{\frac{4}{3}}+2a^{\frac{11}{3}}c^{\frac{4}{3}}+2a^{2}b^{\frac{4}{3}}c^{\frac{4}{3}} \geq 8(a^{\frac{64}{3}}b^8c^8)^{\frac{1}{8}}=8a^{\frac{8}{3}}bc</cmath> | ||
+ | as desired, completing the proof of the claim. | ||
+ | |||
+ | Similarly <math>\frac{b}{\sqrt{b^2+8ca}} \geq \frac{b^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}</math> and <math>\frac{c}{\sqrt{c^2+8ab}} \geq \frac{c^{\frac{4}{3}}}{a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}}</math>. | ||
+ | Summing the three inequalities, we obtain the original inequality. | ||
== See also == | == See also == |
Revision as of 10:56, 22 July 2012
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: but by AMGM, and thus the inequality is proven.
Alternate Solution using Isolated Fudging
We claim that Cross-multiplying, squaring both sides and expanding, we have After cancelling the term, we apply AM-GM to RHS and obtain as desired, completing the proof of the claim.
Similarly and . Summing the three inequalities, we obtain the original inequality.
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 |