## Task 27/80

All real numbers $x$ are to be determined for which the function is minimal

$$
f(x)=\sqrt{x+4 \sqrt{x-4}}+\sqrt{x-4 \sqrt{x-4}}
$$
The identity holds:

$$
x \pm 4 \sqrt{x-4} \equiv (2 \pm \sqrt{x-4})^{2}
$$

Using this, we can write $f(x)$ in the form

$$
f(x)=|2+\sqrt{x-4}|+|2-\sqrt{x-4}|
$$

For $4 \leq x \leq 8$, $\sqrt{x-4} \leq 2$ (for $x > 8$, $\sqrt{x-4} > 2$). This implies

$$
f(x)=2+\sqrt{x-4}-(2-\sqrt{x-4})=2 \sqrt{x-4} > 4
$$

Therefore, $f_{\min}(x)=4$ for all $x$ with $4 \leq x \leq 8$.