Minimize function $\sqrt{x^2-4x+5}+\sqrt{4+x^2}$ Suppose I want to find
$$
\min f(x) = \min(\sqrt{x^2-4x+5}+\sqrt{4+x^2})
$$
I'd start with computing derivative and set it to $0$
$$
f'(x) = \frac{x-2}{\sqrt{x^2-4x+5}} + \frac{x}{\sqrt{4+x^2}}
$$
Then
$$
\frac{x-2}{\sqrt{x^2-4x+5}} + \frac{x}{\sqrt{4+x^2}} = 0
$$
$$
(x-2)\sqrt{4+x^2} + x\sqrt{x^2-4x+5}=0
$$
My first instinct is to rewrite it as
$$
(2-x)\sqrt{4+x^2} = x\sqrt{x^2-4x+5}
$$
and square both sides. The thing is, I get
$$
3x^2 -16x +16=0
$$
with $x_1 = 4/3$ and $x_2 = 4$. So $f'(x_1^-) <0$ and $f'(x_1^+) > 0$ thus in $x = 4/3$ there is a minimum of our function.
Question: is it valid to square both sides of an equation as I did above? Is there a possibility that because of that I lose or introduce a solution that shouldn't exist and might mess everything up?

You can square, but only after ensuring that you're equating numbers that are either both positive or both negative.
In your case you have
$$\textstyle
(2-x)\sqrt{4+x^2}=x\sqrt{x^2-4x+5}
$$
The square roots give no problem: they are both defined anywhere and positive. The sign is determined by the other factors.
If $2-x\ge0$, we also need $x\ge0$, that is, $0\le x\le2$.
If $2-x\le 0$, we also need $x\le0$, but this cannot happen.
Thus we can square, but we must discard roots that are not in $[0,2]$.
The equation you get is indeed $3x^2-16x-16=0$, with roots $4/3$ and $4$. The latter must be discarded.

By the way, the problem can be solved in a different way. Your function is
\begin{align}
f(x)
&=\textstyle\sqrt{x^2+4}+\sqrt{(x-2)^2+1^2} \\[6px]
&=\textstyle\sqrt{(x-0)^2+(0-2)^2}+\sqrt{(x-2)^2+(0-(-1))^2}
\end{align}
which is the sum of the distances of a point on the line $x=0$ from the points $(0,2)$ and $(2,-1)$.
You don't even need to compute the point of minimum, because the minimum is the distance between the two points, hence
$$\textstyle
\sqrt{(0-2)^2+(2+1)^2}=\sqrt{4+9}=\sqrt{13}
$$
If you want to find the point of minimum, the line joining the two points has equation
$$
\frac{y+1}{2+1}=\frac{x-2}{0-2}
$$
The only point $(x,0)$ on the $x$-axis satisfies
$$
\dfrac{1}{3}=\dfrac{2-x}{2}
$$
hence $x=4/3$.

