<s>[INST] 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?
 [/INST] Squaring both sides gives you the solutions to $(2-x)\sqrt{4+x^2} = x\sqrt{x^2-4x+5}$ and to $(2-x)\sqrt{4+x^2} = -x\sqrt{x^2-4x+5}$. So if our possible candidates are $x = 4/3, 4$, we can substitute these values into the equation before squaring.
We find that only $x_1 = 4/3$ satisfies the original equation.
Of course it remains to check that $(x_1, f(x_1))$ is a minimum. Therefore $f(x)$ has exactly one minimum point and no other extrema, as the graph confirms:

</s>