Example 13. Solve the inequality

$$
\sqrt{x^{2}-8 x+15}+\sqrt{x^{2}+2 x-15}>\sqrt{4 x^{2}-18 x+18}
$$
Solution. Since

$$
\begin{gathered}
x^{2}-8 x+15=(x-3)(x-5) \\
x^{2}+2 x-15=(x+5)(x-3) \\
4 x^{2}-18 x+18=4(x-3)(x-3 / 2)
\end{gathered}
$$

the domain of the inequality consists of $x=3, x \leqslant-5$ and $x \geqslant 5$.

For $x=3$, both sides of the inequality (18) are equal to zero; therefore, the number $x=3$ is not a solution to the inequality (18). Thus, the inequality (18) is equivalent to the system

$$
\left\{\begin{array}{l}
|x| \geqslant 5 \\
\sqrt{x^{2}-8 x+15}+\sqrt{x^{2}+2 x-15}>\sqrt{4 x^{2}-18 x+18}
\end{array}\right.
$$

Both sides of the irrational inequality are non-negative; therefore, after squaring and simplifying, we obtain the system

$$
\left\{\begin{array}{l}
|x| \geqslant 5 \\
\sqrt{x^{2}-8 x+15} \sqrt{x^{2}+2 x-15}>(x-3)^{2}
\end{array}\right.
$$

which is equivalent to the last one. This, in turn, is equivalent to the system

i.e., the system

$$
\left\{\begin{array}{l}
|x| \geqslant 5 \\
\sqrt{(x-3)^{2}\left(x^{2}-25\right)}>(x-3)^{2}
\end{array}\right.
$$

$$
\left\{\begin{array}{l}
|x| \geqslant 5 \\
(x-3)^{2}\left(x^{2}-25\right)>(x-3)^{4}
\end{array}\right.
$$

Thus, the inequality (18) is equivalent to the system

$$
\left\{\begin{array}{l}
|x| \geqslant 5 \\
x^{2}-25>(x-3)^{2}
\end{array}\right.
$$

Solving this system, we find that it, and therefore the inequality (18), is satisfied by all numbers from the interval $x>17 / 3$.