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!
 A: Let $f(x)=\sqrt{-x^2+4x+21}+\sqrt{-x^2+3x+10}.$
Thus, $f$ is a concave function, which gives
$$\min_{[-2,5]}f=\min\{f(-2),f(5)\}=f(-2)=3.$$
I used the following obvious statement.

Let $g(x)=\sqrt{f(x)},$ where $f$ is a concave twice differentiable function on $\{x|f(x)\geq0\}.$
Prove that $g$ is a concave function.

A: Let
\begin{align*}
f_1(x)&=\sqrt{-x^2+4x+21}\\
f_2(x)&=\sqrt{-x^2+3x+10}.
\end{align*}
An examination of the domains of each of these functions separately shows that 
\begin{align*}
\mathcal{D}(f_1)&=[-3,7]\\
\mathcal{D}(f_2)&=[-2,5].
\end{align*}
If we define $f(x)=f_1(x)+f_2(x),$ then the above forces the domain $\mathcal{D}(f)=[-2,5].$ Therefore, we must evaluate the function $f$ at $-2$ and $5$. However, there may be critical points inside the domain. Differentiating $f$ yields
$$f'(x)=\frac{2-x}{\sqrt{-x^2+4x+21}}+\frac{3/2-x}{\sqrt{-x^2+3x+10}}. $$
Setting this equal to zero yields the following:
\begin{align*}
(2-x)\sqrt{-x^2+3x+10}&=(x-3/2)\sqrt{-x^2+4x+21}\\
(2-x)^2(-x^2+3x+10)&=(x-3/2)^2(-x^2+4x+21)\\
&\vdots\\
51x^2-104x+29&=0\\
x&=\frac{29}{17},\;\frac{1}{3},
\end{align*}
both of which are in $\mathcal{D}(f).$ However, plugging $1/3$ into $f'(x)$ shows that we picked up a spurious root by squaring. Evaluating, then, we have
\begin{align*}
f(-2)&=3 \\
f\left(\frac{29}{17}\right)&=6\sqrt{2}\approx 8.49 \\
f(5)&=4.
\end{align*}
This proves that the minimum occurs at $x=-2,$ and has a value of $y=3.$
A: 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 $x=-2$.
