<s>[INST] Find minimum of $y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$ 
Find minimum of $y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$

My work so far:
1) $$y =\frac{2x+9}{\sqrt{-x^2+4x+12}+\sqrt{-x^2+2x+3}} $$
2) I used a derivative and found the answer ($y=\sqrt3$ at $x=0$). Is there any other way?
 [/INST] Domain gives $-1\leq x\leq3$.
If $x=0$ so $y=\sqrt3$.
We'll prove that it's a minimal value of $y$.
Id est, we need to prove that 
$$\sqrt{-x^2+2x+3}+\sqrt3\leq\sqrt{-x^2+4x+12}$$ or after squaring of the both sides we need to prove that
$$\sqrt{3(-x^2+2x+3)}\leq x+3$$ or since $x+3>0$, we need to prove that
$$3(-x^2+2x+3)\leq(x+3)^2,$$
which is $x^2\geq0$.
Done!
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