Question #82d1a

1 Answer
Oct 18, 2017

See below.

Explanation:

#log_x y/(x+y)+log_y z/(y+z)+log_z x/(z+x)>=9/(2(x+y+z))#

Calling #f(x,y,z) = log_x y/(x+y)+log_y z/(y+z)+log_z x/(z+x)-9/(2(x+y+z))#

we will demonstrate that #f(x,y,z) ge 0#

By symmetry, (it is symmetric because interchanging any of #x,y,z# the expression does not change) the solution is at

#x=y=z=lambda# and

#3/(2lambda) - 9/(2 xx 3lambda) = 0# or

#f(x,y,z) ge f(lambda,lambda,lambda) ge 0#

and then

#log_x y/(x+y)+log_y z/(y+z)+log_z x/(z+x)>=9/(2(x+y+z))# for all #0 < {x,y,z} < oo# and #{x,y,z} ne {1,1,1}#

NOTE:

We have #grad f(lambda,lambda,lambda) = 0# and

#grad(grad f(lambda,lambda,lambda)) ge 0# which is semi-positive definite.