9.143. $\sqrt{x+3}<\sqrt{x-1}+\sqrt{x-2}$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}x+3 \geq 0, \\ x-1 \geq 0, \\ x-2 \geq 0\end{array} \Leftrightarrow x \geq 2\right.$.

Since both sides of the inequality are non-negative, squaring both sides, we get

$$
\begin{aligned}
& x+36-x . \text { The last inequality, taking into account the domain of definition, is equivalent to the following two systems of inequalities: }
\end{aligned}
$$

1) $\left\{\begin{array}{l}6-x6 \\ (x-1)(x-2) \geq 0\end{array} \Leftrightarrow x>6\right.\right.$
2) $\left\{\begin{array}{l}6-x \geq 0, \\ x \geq 2, \\ 4(x-1)(x-2)>(6-x)^{2}\end{array} \Leftrightarrow\left\{\begin{array}{l}2 \leq x \leq 6, \\ x^{2}>\frac{28}{3}\end{array} \Leftrightarrow\left\{\begin{array}{l}2 \leq x \leq 6, \\ x>\sqrt{\frac{28}{3}}\end{array} \Leftrightarrow \sqrt{\frac{28}{3}}\sqrt{\frac{28}{3}}\right.\right.\right.$.

Answer: $x \in\left(\sqrt{\frac{28}{3}} ; \infty\right)$.