74. Let's solve the inequality

$$
\sqrt{x+3}<\sqrt{x-1}+\sqrt{x-2}
$$
Solution. a) $f(x)=\sqrt{x+3}-\sqrt{x-1}-\sqrt{x-2}$. Let's find the domain of definition:

$$
\left(\{ \begin{array} { l } 
{ x + 3 \geqslant 0 , } \\
{ x - 1 \geqslant 0 , } \\
{ x - 2 \geqslant 0 }
\end{array} ) \Leftrightarrow \left(\left\{\begin{array}{l}
x \geqslant-3, \\
x \geqslant 1, \\
x \geqslant 2
\end{array}\right) \Leftrightarrow(x \geqslant 2), \text { i.e. } D(f)=[2 ;+\infty)\right.\right.
$$

b) Solving the equation $\sqrt{x+3}=\sqrt{x-1}+\sqrt{x-2}$, we get the only root $x=\frac{2}{3} \sqrt{21}$. Therefore, the function $f(x)$ has only one zero: $x=\frac{2}{3} \sqrt{21}$.

c) Mark on the coordinate line (Fig. 41) $x=\frac{2}{3} \sqrt{21}$.

If $x \in\left[2 ; \frac{2}{3} \sqrt{21}\right)$, then $f(x)>0$; if $x \in\left(\frac{2}{3} \sqrt{21} ;+\infty\right)$, then $f(x)<0$. Answer: $\left(\frac{2}{3} \sqrt{21} ;+\infty\right)$.