SOLUTION: show that ((x^2y)/2(x+y))+ ((x+y)/2xy)-(x)^1/2 >= 0 for positive real numbers x and y.

Question 28086: show that ((x^2y)/2(x+y))+ ((x+y)/2xy)-(x)^1/2 >= 0 for positive real numbers x and y.
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Prove this is true for positive real x & y

I think this is the right approach - I admit I'm a little over my head
If I multiply both sides by 2, the equality stll holds true.

Now if I add to both sides, the equality is still true.

I can rewrite the second term by separating into two fractions
=

Now I choose to multiply both top and bottom of the first term by

Now I think I have to let x and y be 0 or very large in all possible combinations
(a) x = 0 and y = 0
(b) x = 0 and y approaches infinity
(c) x approaches infinity and y = 0
(d) x approaches infinity and y approaches infinity
That covers the extremes of all real values that x in combination with y can have
(a) the first term is 0/infinity = 0 and the other terms = infinity
so this satisfies the equation
(b) the first term is still 0/infinity
the other terms are 0 + infinity = infinity, so the equation still holds
(c) The first term approaches infinity, equation holds
(d) The first term approaches infinity, equation holds
Hope this is correct and helps
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