Finding the minimum value of a rational function. Prove that if $x$ is real and $a>c$ & $b>c$ the minimum value of $$\frac{(a+x)(b+x)}{(c+x)} ;Given\space( x>-c)$$ is $$({\sqrt{a-c}+\sqrt{b-c \space}})^2$$
I tried using minima condition but the expression was too complicated to arrange and solve .How can this be solved algebraically? Please give answers/suggestions. 
Thanks.

Expand and simplify (or rather, do partial fraction decomposition), we get that
$$ \frac{ (a+x)(b+x)}{c+x} = (x+c) + \frac{(a-c)(b-c)}{x+c} + a+b-2c $$
Apply AM-GM to the first two terms to conclude that
$$ (x+c) + \frac{(a-c)(b-c)}{x+c} + a+b-2c \geq 2 \sqrt{ (a-c)(b-c)  } + a+b-2c = \left( \sqrt{a-c} + \sqrt{b-c} \right)^2$$
