Problem A. 561. (April 2012)
A. 561. Show that
holds for all positive numbers a, b, c, p.
Deadline expired on 10 May 2012.
Solution 1. We show that
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.
Lemma. For A,p>0 we have
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.