find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$ Question:
Given $x + y = 12$, find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$?
Key:
I use $y = 12 - x$ and substitute into the equation, and derivative it.
which I got this 
$$\frac{x}{\sqrt{x^2+4}} + \frac{x-12}{\sqrt{x^2-24x+153}} = f'(x).$$
However, after that. I don't know how to do next in order to find the minimum value. Please help! 

Setting $f'(x)=0$ gives 
$$\frac{x}{\sqrt{x+4}}+\frac{x-12}{\sqrt{x^2-24x+153}}=0 $$
or
$$\frac{x}{\sqrt{x+4}}=-\frac{x-12}{\sqrt{x^2-24x+153}}\tag1$$
Squaring both sides of $(1)$ gives
$$\left(\frac{x}{\sqrt{x^2+4}}\right)^2=\left(\frac{x-12}{\sqrt{x^2-24x+153}}\right)^2$$
which can be rewritten as
$$\frac{x^2}{x^2+4}=\frac{(x-12)^2}{x^2-24x+153} \tag2$$
Thus, writing the numerator of the left-hand side as $x^2=x^2+4-4$ and writing the numerator of the right-hand side as $(x-12)^2=(x-12)^2+9-9$ simplifies to $(2)$
$$\frac{4}{x^2+4}=\frac{9}{x^2-24x+153} \tag3$$
Can you finish from here?
