Difference between revisions of "1995 IMO Problems/Problem 2"
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=== Solution 3b === | === Solution 3b === | ||
Without clever notation: | 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> | + | By Cauchy-Schwarz, <cmath>\left(a(b+c) + b(c+a) + c(a+b)\right) \cdot \left(\frac{1}{a^3 (b+c)} + \frac{1}{b^3 (c+a)} + \frac{1}{c^3 (a+b)}\right)</cmath> |
− | <cmath>\ge (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^2</cmath> | + | <cmath>\ge \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)^2</cmath> |
<cmath>= (ab + bc + ac)^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 | + | Dividing by <math>2(ab + bc + ac)</math> 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> | <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 08:27, 15 July 2017
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
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.