# 3. Solution:

Domain of Definition (ODZ): $\left\{\begin{array}{c}x^{2}-4 x+3 \geq 0 \\ 3-x \geq 0 \\ x^{2}-6 x+9 \geq 0 \\ x^{2}-8 x+16 \geq 0 \\ x^{2}+2 \cdot \sqrt{x^{2}-8 x+16} \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{c}(x-1) \cdot(x-3) \geq 0 \\ x \leq 3 \\ (x-3)^{2} \geq 0 \\ (x-4)^{2} \geq 0 \\ x^{2}+2 \cdot \sqrt{(x-4)^{2}} \neq 0\end{array} \Leftrightarrow\right.\right.$ $x \in(-\infty ; 1] \cup\{3\}$.

On the ODZ, $\sqrt{x^{2}-6 x+9}=\sqrt{(x-3)^{2}}=|x-3|=3-x$;

$\sqrt{x^{2}-8 x+16}=\sqrt{(x-4)^{2}}=|x-4|=4-x$. We obtained:

$\frac{(\sqrt{(x-1) \cdot(x-3)}+\sqrt{3-x})^{2} \cdot\left(x^{2}-4(3-x)\right)}{x^{2}+2(4-x)} \geq 0 \Leftrightarrow \frac{(\sqrt{(x-1) \cdot(x-3)}+\sqrt{3-x})^{2} \cdot(x+6) \cdot(x-2)}{x^{2}-2 x+8} \geq 0 \Leftrightarrow$

$x \in(-\infty ;-6] \cup\{3\}$.
Answer: $(-\infty ;-6] \cup\{3\}$.