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

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: show that ((x^2y)/2(x+y))+ ((x+y)/2xy)-(x)^1/2 >= 0 for positive real numbers x and y.      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
Prove this is true for positive real x & y
%28%28x%5E2y%29%2F2%28x%2By%29%29%2B+%28%28x%2By%29%2F2xy%29-%28x%29%5E%281%2F2%29+%3E=+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.
%28%28x%5E2y%29%2F%28x%2By%29%29%2B+%28%28x%2By%29%2Fxy%29-2%2A%28x%29%5E%281%2F2%29+%3E=+0
Now if I add 2%2A+%28x%29%5E%281%2F2%29 to both sides, the equality is still true.
%28%28x%5E2y%29%2F%28x%2By%29%29%2B+%28%28x%2By%29%2Fxy%29+%3E=+2%2A%28x%29%5E%281%2F2%29
I can rewrite the second term by separating into two fractions
%28x%2By%29%2Fxy = 1%2Fy+%2B+1%2Fx
%28%28x%5E2y%29%2F%28x%2By%29%29%2B+1%2Fy+%2B+1%2Fx+%3E=+2%2A%28x%29%5E%281%2F2%29
Now I choose to multiply both top and bottom of the first term by 1%2F%28x%2Ay%29
+%28x+%2F%281%2Fy+%2B+1%2Fx%29%29+%2B+1%2Fy+%2B+1%2Fx+%3E=+2%2A%28x%29%5E%281%2F2%29
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