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A. 561. Show that

holds for all positive numbers a, b, cp.

(5 points)

Deadline expired on 10 May 2012.

Solution 1. We show that

 (1)

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

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

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

Solution 2.

Lemma. For A,p>0 we have

where

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

Proof. Substituting u=At,

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,

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

Statistics on problem A. 561.
 5 students sent a solution. 5 points: Ágoston Tamás, Janzer Olivér, Mester Márton, Omer Cerrahoglu, Strenner Péter.

• Problems in Mathematics of KöMaL, April 2012

•  Támogatóink: Morgan Stanley