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.
|Statistics on problem A. 561.|
Problems in Mathematics of KöMaL, April 2012