Difference between revisions of "1995 IMO Problems/Problem 2"
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− | Note that <math>abc = 1 \implies a = \frac{1}{bc}</math>. The cyclic sum becomes <math>\sum_{cyc}\frac{(bc)^3}{b + c}</math>. Note that by AM-GM, the cyclic sum is greater than or equal to <math>3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^{\frac13}</math>. We now see that we have the three so we must be on the right path. We now only need to show that <math>\frac32 \geq 3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^\frac13</math>. Notice that by AM-GM, <math>a + b \geq 2\sqrt{ab}</math>, <math>b + c \geq 2\sqrt{bc}</math>, and <math>a + c \geq 2\sqrt{ac}</math>. Thus, we see that <math>(a+b)(b+c)(a+c) \geq 8</math>, concluding that <math>\sum_{cyc} \frac{(bc)^3}{b+c} \geq \frac32 \geq 3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^{\frac13}</math> | + | Note that <math>abc = 1 \implies a = \frac{1}{bc}</math>. The cyclic sum becomes <math>\sum_{cyc}\frac{(bc)^3}{b + c}</math>. Note that by AM-GM, the cyclic sum is greater than or equal to <math>3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^{\frac13}</math>. We now see that we have the three so we must be on the right path. We now only need to show that <math>\frac32 \geq 3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^\frac13</math>. Notice that by AM-GM, <math>a + b \geq 2\sqrt{ab}</math>, <math>b + c \geq 2\sqrt{bc}</math>, and <math>a + c \geq 2\sqrt{ac}</math>. Thus, we see that <math>(a+b)(b+c)(a+c) \geq 8</math>, concluding that <math>\sum_{cyc} \frac{(bc)^3}{b+c} \geq \frac32 \geq 3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^{\frac13}</math>. |
+ | ^ This solution is incorrect, as it does not prove inequalities in the right direction. Proving that <math>A \geq B</math>, and <math>C \geq B</math> does not show that <math>A \geq C \geq B</math>. | ||
=== Solution 7 from Brilliant Wiki (Muirheads) ==== | === Solution 7 from Brilliant Wiki (Muirheads) ==== | ||
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[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||
+ | {{IMO box|year=1995|num-b=1|num-a=3}} |
Latest revision as of 00:01, 26 July 2024
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 .
^ This solution is incorrect, as it does not prove inequalities in the right direction. Proving that , and does not show that .
Solution 7 from Brilliant Wiki (Muirheads) =
https://brilliant.org/wiki/muirhead-inequality/
Solution 8 (fast Titu's Lemma no substitutions)
Rewrite as .
Now applying Titu's lemma yields .
Now applying the AM-GM inequality on . The result now follows.
Note: , because . (Why? Because , and hence ).
~th1nq3r
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Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
Resources
1995 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |