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Problem A. 561. (April 2012)

A. 561. Show that


\frac{a^3b}{{(3a+b)}^p} + \frac{b^3c}{{(3b+c)}^p}
+\frac{c^3a}{{(3c+a)}^p} \ge \frac{a^2bc}{{(2a+b+c)}^p} +
\frac{b^2ca}{{(2b+c+a)}^p} + \frac{c^2ab}{{(2c+a+b)}^p}

holds for all positive numbers a, b, cp.

(5 pont)

Deadline expired on 10 May 2012.


Solution 1. We show that


\frac{a^3b}{{(3a+b)}^p} + \frac{c^2ab}{{(2c+a+b)}^p}
\ge \frac{2a^2bc}{{(2a+b+c)}^p}. (1)

Applying the AM-GM inequality to the fractions on the left-hand side, then to the numbers 3a+b and a+b+2c,


\frac{a^3b}{{(3a+b)}^p} + \frac{c^2ab}{{(2c+a+b)}^p} \ge
2\sqrt{\frac{a^3b}{{(3a+b)}^p} \cdot \frac{c^2ab}{{(2c+a+b)}^p}} =

 = \frac{2a^2bc}{\Big(\sqrt{(3a+b)(a+b+2c)}\Big)^p} \ge
\frac{2a^2bc}{\Big(\frac{(3a+b)+(a+b+2c)}2\Big)^p} =
\frac{2a^2bc}{(2a+b+c)^p}.

Rearranging (1), by the cyclic permutations of the variables we get


\frac{a^3b}{{(3a+b)}^p} \ge
2\frac{a^2bc}{{(2a+b+c)}^p} - \frac{c^2ab}{{(2c+a+b)}^p},


\frac{b^3c}{{(3b+c)}^p} \ge
2\frac{b^2ca}{{(2b+c+a)}^p} - \frac{a^2bc}{{(2a+b+c)}^p},


\frac{c^3a}{{(3c+a)}^p} \ge
2\frac{c^2ab}{{(2c+a+b)}^p} - \frac{b^2ca}{{(2b+c+a)}^p}.

The statement of the problem is the sum of the last three lines above.

Solution 2.

Lemma. For A,p>0 we have


\frac1{A^p} = \frac1{\Gamma(p)} \int_0^\infty
t^{p-1}e^{-At}\mathrm{d}t,

where


\Gamma(p) = \int_0^\infty t^{p-1}e^{-t}\mathrm{d}t

is Euler's Gamma-function. (See also the solution of Problem A. 493.)

Proof. Substituting u=At,


\int_0^\infty t^{p-1}e^{-At} \, \mathrm{d}t =
\frac1{A^p}
\int_0^\infty u^{p-1}e^{-u} \, \mathrm{d}u =
\frac{\Gamma(p)}{A^p}.

Applying the Lemma to A=3a+b, A=3b+c, A=3c+a, A=2a+b+c, A=2b+c+a, A=2c+a+b,


\left(\frac{a^3b}{{(3a+b)}^p} + \frac{b^3c}{{(3b+c)}^p}
+\frac{c^3a}{{(3c+a)}^p}\right) - \left(\frac{a^2bc}{{(2a+b+c)}^p}+
\frac{b^2ca}{{(2b+c+a)}^p} + \frac{c^2ab}{{(2c+a+b)}^p} \right) =


=\frac{a^3b}{\Gamma(p)}\int_0^\infty t^{p-1}e^{-(3a+b)t}\mathrm{d}t 
+\frac{b^3c}{\Gamma(p)}\int_0^\infty t^{p-1}e^{-(3b+c)t}\mathrm{d}t 
+\frac{c^3a}{\Gamma(p)}\int_0^\infty t^{p-1}e^{-(3c+a)t}\mathrm{d}t =


-\frac{a^2bc}{\Gamma(p)}\int_0^\infty t^{p-1}e^{-(2a+b+c)t}\mathrm{d}t 
-\frac{b^2ca}{\Gamma(p)}\int_0^\infty t^{p-1}e^{-(2b+c+a)t}\mathrm{d}t 
-\frac{c^2ab}{\Gamma(p)}\int_0^\infty t^{p-1}e^{-(2c+a+b)t}\mathrm{d}t


=\frac1{\Gamma(p)}\int_0^\infty t^{p-1} \Big(
a^3b e^{-(3a+b)t} +
b^3c e^{-(3b+c)t} +
c^3a e^{-(3c+a)t} -


- a^2bc e^{-(2a+b+c)t} -
b^2ca e^{-(2b+c+a)t} -
c^2ab e^{-(2c+a+b)t}
\Big)\mathrm{d}t


=\frac1{\Gamma(p)}\int_0^\infty t^{p-1} \bigg(
abe^{-(a+b)t}\Big(ae^{-at}-ce^{-ct}\Big)^2 +
bce^{-(b+c)t}\Big(be^{-bt}-ae^{-at}\Big)^2 +


+ cae^{-(c+a)t}\Big(ce^{-ct}-be^{-bt}\Big)^2 
\bigg)\mathrm{d}t \ge0.

Remark. Both soutions show that equality holds only when a=b=c.


Statistics:

5 students sent a solution.
5 points:Ágoston Tamás, Janzer Olivér, Mester Márton, Omer Cerrahoglu, Strenner Péter.

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