Question 28086
Prove this is true for positive real x & y
{{{((x^2y)/2(x+y))+ ((x+y)/2xy)-(x)^(1/2) >= 0}}}
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.
{{{((x^2y)/(x+y))+ ((x+y)/xy)-2*(x)^(1/2) >= 0}}}
Now if I add {{{2* (x)^(1/2)}}} to both sides, the equality is still true.
{{{((x^2y)/(x+y))+ ((x+y)/xy) >= 2*(x)^(1/2)}}}
I can rewrite the second term by separating into two fractions
{{{(x+y)/xy}}} = {{{1/y + 1/x}}}
{{{((x^2y)/(x+y))+ 1/y + 1/x >= 2*(x)^(1/2)}}}
Now I choose to multiply both top and bottom of the first term by {{{1/(x*y)}}}
{{{ (x /(1/y + 1/x)) + 1/y + 1/x >= 2*(x)^(1/2)}}}
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