Difference between revisions of "1995 IMO Problems/Problem 2"
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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 | 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> | + | <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. | as desired. | ||
Revision as of 22:18, 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.