Difference between revisions of "1995 IMO Problems/Problem 2"
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By Cauchy-Schwarz, <cmath>\left(\sum_{cyc}\dfrac{1}{a^3 (b+c)}\right)\left(\sum_{cyc}a(b+c)\right)\geq \left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right) ^2=(ab+ac+bc)^2</cmath> Dividing by <math>2(ab+bc+ac)</math> gives <cmath>\dfrac{1}{a^3 (b+c)}+\dfrac{1}{b^3 (a+c)}+\dfrac{1}{c^3 (a+b)}\geq \dfrac{1}{2}(ab+bc+ac)\geq \dfrac{3}{2}</cmath> by AM-GM. | By Cauchy-Schwarz, <cmath>\left(\sum_{cyc}\dfrac{1}{a^3 (b+c)}\right)\left(\sum_{cyc}a(b+c)\right)\geq \left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right) ^2=(ab+ac+bc)^2</cmath> Dividing by <math>2(ab+bc+ac)</math> gives <cmath>\dfrac{1}{a^3 (b+c)}+\dfrac{1}{b^3 (a+c)}+\dfrac{1}{c^3 (a+b)}\geq \dfrac{1}{2}(ab+bc+ac)\geq \dfrac{3}{2}</cmath> by AM-GM. | ||
+ | |||
+ | === Solution 3b === | ||
+ | Without clever notation: | ||
+ | By Cauchy-Schwarz, <cmath>(a(b+c) + b(c+a) + c(a+b)) \cdot (\frac{1}{a^3 (b+c)} + \frac{1}{b^3 (c+a)} + \frac{1}{c^3 (a+b)})</cmath> | ||
+ | <cmath>\ge (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^2</cmath> | ||
+ | <cmath>= (ab + bc + ac)^2</cmath> | ||
+ | |||
+ | Dividing by 2(ab + bc + ac) and noting that <math>ab + bc + ac \ge 3(a^2b^2c^2)^{\frac{1}{3}} = 3</math> by AM-GM gives | ||
+ | <cmath>(frac{1}{a^3 (b+c)} + \frac{1}{b^3 (c+a)} + \frac{1}{c^3 (a+b)} \ge \frac{ab + bc + ac}{2} \ge \frac{3}{2},</cmath> | ||
+ | as desired. | ||
=== Solution 4 === | === Solution 4 === |
Revision as of 22:17, 21 April 2014
Contents
[hide]Problem
(Nazar Agakhanov, Russia) Let be positive real numbers such that . Prove that
Solution
Solution 1
We make the substitution , , . Then Since and are similarly sorted sequences, it follows from the Rearrangement Inequality that By the Power Mean Inequality, Symmetric application of this argument yields Finally, AM-GM gives us as desired.
Solution 2
We make the same substitution as in the first solution. We note that in general, It follows that and are similarly sorted sequences. Then by Chebyshev's Inequality, By AM-GM, , and by Nesbitt's Inequality, The desired conclusion follows.
Solution 3
Without clever substitutions: By Cauchy-Schwarz, Dividing by gives by AM-GM.
Solution 3b
Without clever notation: By Cauchy-Schwarz,
Dividing by 2(ab + bc + ac) and noting that by AM-GM gives as desired.
Solution 4
Proceed as in Solution 1, to arrive at the equivalent inequality But we know that by AM-GM. Furthermore, by Cauchy-Schwarz, and so dividing by gives as desired.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.