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.
Your last equation is correct. Solving it for $x$ gives you the two extrema of your function; say that the roots of the quadratic equation (your last one) are $x_1$ and $x_2$. Because of the signs, $x_1$ corresponds to the maximum and $x_2$ to the minimum of the function. Compute now the corresponding value $y_2$ (you will need to work for simplifying them). From what I got, $$y_2 = (a + b - 2 c) + 2 \sqrt{(a-c) (b-c)}$$ Manipulating $y_2$ shows that it is equal to $$(\sqrt{a-c} + \sqrt{b-c})^2$$ I hope and wish this helps you to continue.