<s>[INST] Find the minimum value of $y = \sqrt {{x^2} + 4x + 13} + \sqrt {{x^2} - 8x + 41} $ Find the minimum value of $y = \sqrt {{x^2} + 4x + 13}  + \sqrt {{x^2} - 8x + 41} $ .
My approach is as follow
$y = \sqrt {{x^2} + 4x + 4 + 9}  + \sqrt {{x^2} - 8x + 16 + 25}  \Rightarrow y = \sqrt {{{\left( {x + 2} \right)}^2} + 9}  + \sqrt {{{\left( {x - 4} \right)}^2} + 25} $
Within the square roots the minimum values are 3 and 5 hence the minimum value should be greeter than 8 but not able to find the actual minimum value.
 [/INST] I was going to do a geometrical solution, but looks like that's already been done.
As for a calculus solution, we have
$$\frac{dy}{dx}=\frac{x+2}{\sqrt{x^2+4x+13}}+\frac{x-4}{\sqrt{x^2-8x+41}}$$
As, you have already shown $\sqrt{x^2+4x+13}>0$ and $\sqrt{x^2-8x+41}>0$ from completing the square. Hence, the only critical points occur when the RHS is equal to zero.
$$\frac{x+2}{\sqrt{x^2+4x+13}}+\frac{x-4}{\sqrt{x^2-8x+41}}=0$$
$$\frac{x+2}{\sqrt{x^2+4x+13}}=-\frac{x-4}{\sqrt{x^2-8x+41}}$$
$$\frac{x^2+4x+4}{x^2+4x+13}=\frac{x^2-8x+16}{x^2-8x+41}$$
$$\frac{x^2+4x+4}{x^2+4x+13}-1=\frac{x^2-8x+16}{x^2-8x+41}-1$$
$$\frac{-9}{x^2+4x+13}=\frac{-25}{x^2-8x+41}$$
$$9x^2-72x+369=25x^2+100x+325$$
$$16x^2+172x-44=0$$
$$4x^2+43x-11=0$$
$$(4x-1)(x+11)=0$$
$$x=-11,\frac{1}{4}$$
However, since we squared both sides of the equation, we expect one of the solutions to be extranneous. Note that if $x=-11$, then $x+2$ and $x-4$ are both negative. Hence, $\frac{x+2}{\sqrt{x^2+4x+13}}$ and $\frac{x-4}{\sqrt{x^2-8x+41}}$ will have the same sign. Then of course we could not have $\frac{x+2}{\sqrt{x^2+4x+13}}=-\frac{x-4}{\sqrt{x^2-8x+41}}$. So our solution is $x=\frac{1}{4}$. It is not hard to deduce that this is a minima.
So the minimum value of $y$ is
$$\sqrt{\frac{1}{16}+14}+\sqrt{\frac{1}{16}+39}$$
$$=\sqrt{\frac{225}{16}}+\sqrt{\frac{625}{16}}$$
$$=\frac{15}{4}+\frac{25}{4}$$
$$=\boxed{10}$$
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