Example 22. Solve the inequality

$$
\left(\sqrt{x^{2}-4 x+3}+1\right) \log _{2} \frac{x}{5}+\frac{1}{x}\left(\sqrt{8 x-2 x^{2}-6}+1\right) \leqslant 0
$$
Solution. The domain of admissible values of the original inequality consists of all $x$ satisfying the system

$$
\left\{\begin{array}{l}
x>0 \\
x^{2}-4 x+3 \geqslant 0 \\
8 x-2 x^{2}-6 \geqslant 0
\end{array}\right.
$$

i.e., the system

$$
\left\{\begin{array}{l}
x>0 \\
x^{2}-4 x+3 \geqslant 0 \\
x^{2}-4 x+3 \leqslant 0
\end{array}\right.
$$

It follows that the domain of admissible values consists of all $x$ satisfying the system

$$
\left\{\begin{array}{l}
x>0 \\
x^{2}-4 x+3=0
\end{array}\right.
$$

Since the quadratic equation $x^{2}-4 x+3=0$ has positive roots $x_{1}=1$ and $x_{2}=3$, the domain of admissible values of the original inequality is $x=1$ and $x=3$. The solutions of the original inequality lie within its domain of admissible values; therefore, the solutions are among the numbers 1 and 3.

Let $x=1$. Substituting this value into the left-hand side of inequality (18), we get $\log _{5} \frac{1}{5}+1=-1+1=0$, i.e., the value $x=1$ is a solution of the inequality.

Let $x=3$. Then the left-hand side of the original inequality is

$$
\log _{5} \frac{3}{5}+\frac{1}{3}=\log _{5} 3-1+\frac{1}{3}=\log _{5} 3-\frac{2}{3}=\log _{5} \frac{3}{5^{2 / 3}}
$$

Since $27>25$, then $3>5^{2 / 3}$, i.e., $\frac{3}{5^{2 / 3}}>1$; therefore, $\log _{5} \frac{3}{5^{2 / 3}}>0$. Hence, the value $x=3$ is not a solution of the original inequality.

Thus, the set of solutions of the original inequality consists of the single number $x=1$.