<s>[INST] Find the minimum of the function $y=\sqrt{-x^2+4x+21}+\sqrt{-x^2+3x+10}.$ Find the minimum of the function $$y=\sqrt{-x^2+4x+21}+\sqrt{-x^2+3x+10}.$$

By computer I found $\min_y=3;$ then I will prove $y\ge 3$. After squaring we got $$(x+2)(178-37x)\ge 0\quad \forall -2\le x\le 5.$$
My idea is not beautiful so I need some other solution. Thanks!
 [/INST] Here is a simple demonstration -- not using calculus or concavity arguments --
that the minimum of $y=\sqrt{-x^2+4x+21}+\sqrt{-x^2+3x+10}$ is $3$.
Note that $y=\sqrt{-(x-7)(x+3)}+\sqrt{-(x-5)(x+2)}=\sqrt{5^2-(x-2)^2}+\sqrt{\left(\dfrac72\right)^2-\left(x-\dfrac32\right)^2}.$
So $y$ is defined when $-2\le x\le5$ or $\left(x-\dfrac32\right)^2\le\left(\dfrac72\right)^2.  $ 
When that is the case, $-2\le x\le 6$ or $\left(x- 2\right)^2\le4^2.  $ 
Therefore, $y= \sqrt{5^2-(x-2)^2}+\sqrt{\left(\dfrac72\right)^2-\left(x-\dfrac32\right)^2}\ge\sqrt{5^2-(x-2)^2}\ge\sqrt{5^2-4^2}=3,$
and note that $y=3$ when $x=-2$.
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