Difference between revisions of "1995 IMO Problems/Problem 2"
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=== Solution 8 (fast Titu's Lemma) === | === Solution 8 (fast Titu's Lemma) === | ||
− | Rewrite <math>\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)}</math> as <math>\frac( | + | Rewrite <math>\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)}</math> as <math>\frac{(1/a)^2}{a(b+c)} + \frac{(1/b)^2}{b(a+c)} + \frac{(1/c)^2}{c(a+b)}</math>. Now applying Titu's lemma yields <math>\frac{(1/a)^2}{a(b+c)} + \frac{(1/b)^2}){b(a+c)} + \frac{(1/c)^2}{c(a+b)} \geq \frac{(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^2}{a(b+c) + b(a+c) + c(a+b)} = \frac{(ab + bc + ca)^2}{2(ab + bc + ca)} = \frac{ab + bc + ca}{2}</math>. Now applying the AM-GM inequality on <math>ab + bc +ca \geq 3(((ab)(bc)(ca))^2)^{\frac{1}{3}} = 3</math>. The result now follows. |
~th1nq3r | ~th1nq3r | ||
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Scroll all the way down | Scroll all the way down | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 11:08, 6 January 2022
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 and noting that by AM-GM gives as desired.
Solution 4
After the setting and as so concluding
By Titu Lemma, Now by AM-GM we know that and which concludes to
Therefore we get
Hence our claim is proved ~~ Aritra12
Solution 5
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.
Solution 6
Without clever substitutions, and only AM-GM!
Note that . The cyclic sum becomes . Note that by AM-GM, the cyclic sum is greater than or equal to . We now see that we have the three so we must be on the right path. We now only need to show that . Notice that by AM-GM, , , and . Thus, we see that , concluding that
Solution 7 from Brilliant Wiki (Muirheads) =
https://brilliant.org/wiki/muirhead-inequality/
Solution 8 (fast Titu's Lemma)
Rewrite as . Now applying Titu's lemma yields . Now applying the AM-GM inequality on . The result now follows.
~th1nq3r
Scroll all the way down Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.