Finding the minimum value of a rational function.

Question

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.

Answer 1

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

$$ (x+c) + \frac{(a-c)(b-c)}{x+c} + a+b-2c \geq 2 \sqrt{ (a-c)(b-c) } + a+b-2c = \left( \sqrt{a-c} + \sqrt{b-c} \right)^2$$

Answer 2

Hint: Rewrite the top as $(a-c+c+x)(b-c+c+x)$ and let $y=c+x$. We end up with something of shape $\frac{(p+y)(q+y)}{y}$, where $p$ and $q$ are positive.

You will end up needing to minimize $\frac{pq}{y}+y$, which is probably familiar.