<s>[INST] 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! 
 [/INST] You probably know about setting the derivative equal to $0$. The complexity of the equation we get may be discouraging. Your derivative is
$$\frac{x}{\sqrt{x^2+4}}-\frac{12-x}{\sqrt{x^2-24x+153}}.$$
I would rather write it as
$$\frac{x}{\sqrt{x^2+4}}-\frac{y}{\sqrt{y^2+9}}.$$
Nicer! Set this equal to $0$. So we get
$$\frac{x}{\sqrt{x^2+4}}=\frac{y}{\sqrt{y^2+9}}.$$
Square both sides and simplify a bit (cross-multiply). We get
$$x^2(y^2+9)=y^2(x^2+4).$$
There is some nice cancellation, and we get $9x^2=4y^2$. The rest should not be difficult.
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