Problem 10.1. Solve the inequality

$$
\sqrt{x^{2}-1}+\sqrt{2 x^{2}-3}+x \sqrt{3}>0
$$

Peter Boyvalenkov
10.1. The inequality makes sense for $x \in\left(-\infty,-\frac{\sqrt{6}}{2}\right] \cup\left[\frac{\sqrt{6}}{2},+\infty\right)$. All $x$ in the second interval are solutions of the inequality. Let $x \in\left(-\infty,-\frac{\sqrt{6}}{2}\right]$. Then the inequality is equivalent to

$$
\sqrt{\left(x^{2}-1\right)\left(2 x^{2}-3\right)}>2 \Longleftrightarrow 2 x^{4}-5 x^{2}-1>0
$$

Solving the last inequality gives $x \in\left(-\infty,-\frac{\sqrt{5+\sqrt{33}}}{2}\right)$.

So, the solutions of the given inequality are

$$
x \in\left(-\infty,-\frac{\sqrt{5+\sqrt{33}}}{2}\right) \cup\left[\frac{\sqrt{6}}{2},+\infty\right)
$$