Proving the minimum value of (x+a)(x+b)/(x+c) Show that the minimum value of $\frac {(x+a)(x+b)}{(x+c)}$, where a$\gt$c, b$\gt$c, is $(\sqrt{a-c}+\sqrt{b-c})^{2}$ for real values of x$\gt-c$.
I did $$\frac {(x+a)(x+b)}{(x+c)}=y$$ and then took its discriminant greater than zero. 
This led me to $$y^2-2(a+b-2c)y+(a-b)^2\gt0$$
I also tried differentiating the expression as follows.
$$y'= \frac {(x+c)[2x+(a+b)]-[x^2+(a+b)x+ab]}{(x+c)^2}=0$$
$$\therefore x^2+2cx+(a+b)c-ab=0$$
I am unable to proceed after this. Please help.

I'm going to avoid using calculus to solve this question. Instead, I have a different approach which uses simple algebra.
Let's call this given expression $z$.
Put $y = c + x$
$$⇒z = \frac {(y-c+a)(y-c+b)}{y}$$
Simplifying this by opening the brackets we get,
$$⇒z = \frac {(a-c)(b-c) + y(a-c) + y(b-c) + y^2}{y}$$
$$⇒z = \frac {(a-c)(b-c)}{y} + (a-c) + (b-c) + y$$
Now, here we do a bit of manipulation. We add and subtract $2 \sqrt{(a-c) (b-c)}$. Why? Keep looking.
$$⇒z = \frac {(a-c)(b-c)}{y} + (a-c) + (b-c) + y + 2 \sqrt{(a-c) (b-c)} - 2 \sqrt{(a-c) (b-c)}$$
Here, we club $\frac {(a-c)(b-c)}{y}, y$ and $2 \sqrt{(a-c) (b-c)}$ to get a squared term. That's why I added and subtacted $2 \sqrt{(a-c) (b-c)}$.
$$⇒z = \frac {(a-c)(b-c)}{y} + y + 2 \sqrt{(a-c) (b-c)} + (a-c) + (b-c) - 2 \sqrt{(a-c) (b-c)}$$
$$⇒ z = (\frac {\sqrt{(a-c)(b-c)}}{\sqrt{y}} - \sqrt{y})^2+ (a+b-2c) - 2 \sqrt{(a-c) (b-c)}$$
Now, since $$⇒(\frac {\sqrt{(a-c)(b-c)}}{\sqrt{y}} - \sqrt{y})^2 ≥ 0$$
This is because it is a squared term. It can never be negative, it is only either positive or at the minimum, zero.
So,the expression $z$ is minimum only when  $$⇒(\frac {\sqrt{(a-c)(b-c)}}{\sqrt{y}} - \sqrt{y})^2 = 0$$
Hence, the minimum value of $z$ is, i.e, z reduces to 
$$⇒ z = (a+b-2c) - 2 \sqrt{(a-c) (b-c)}$$
Which upon simplification is,
$$⇒ z = (\sqrt{a-c} + \sqrt{b-c})^2$$
Which is the required minimum value.
