Suppose x,y are positive real numbers. Show that #(x^2-y^2)(1/y - 1/x) >=0#?

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Answer 1 · 2017-05-12T18:13:57

#(x^2-y^2)(1/y-1/x) = (x+y)(x-y)((x-y)/(xy)) = ((x+y)(x-y)^2)/(xy)#

As #x>0, y>0#, clearly:

#1/(xy) > 0#

#(x+y) > 0#

and also:

#(x-y)^2 > 0#

because its the square of a real number.

Then:

#((x+y)(x-y)^2)/(xy) > 0#

because it is the product of three positive quantities.