Example 3. Solve the inequality $x+\sqrt{x+1}>\frac{11}{4}$.
Solution 1 The set of allowed values for the unknown $x$ is $x \geqslant-1$. The original inequality is equivalent to the following inequality:
$$
\sqrt{\mathrm{x}+1}>\frac{11}{4}-\mathrm{x} \text {. }
$$

This inequality is also equivalent to the following two systems of inequalities:
$$
\begin{array}{l}
\text { (1) }\left\{\begin{array} { l } 
{ x \geqslant - 1 , } \\
{ \frac { 1 1 } { 4 } - x < 0 }
\end{array}\right. \\
\Longrightarrow x>\frac{11}{4} ; \\
x>-1,
\end{array}
$$
(2) $\frac{11}{4}-x \geqslant 0$,
$$
\begin{array}{l}
(\sqrt{x+1})^{2}>\left(\frac{11}{4}-x\right)^{2} \\
\Rightarrow\left\{\begin{array}{l}
-1\leqslant x \leqslant \frac{11}{4}, \\
x+1>\left(\frac{11}{4}-x\right)^{2}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
-1\leqslant x \leqslant \frac{11}{4}, \\
x+1>\frac{121}{16}-\frac{11}{2}x+x^2
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
-1\leqslant x \leqslant \frac{11}{4}, \\
0>x^2-\frac{13}{2}x+\frac{105}{16}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
-1\leqslant x \leqslant \frac{11}{4}, \\
\frac{5}{4}<x<\frac{21}{4}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
-1\leqslant x \leqslant \frac{11}{4}, \\
\frac{5}{4}<x
\end{array}\right. \\
\Rightarrow x>\frac{5}{4}
\end{array}
$$

Solution 2 The original inequality is equivalent to the system of inequalities:
$$
\begin{array}{l}
\left\{\begin{array}{l}
x+1 \geqslant 0, \\
\left(\sqrt{x+1}+\frac{1}{2}\right)^{2}-\frac{5}{4}>\frac{11}{4}
\end{array}\right. \\
\Longrightarrow\left\{\begin{array}{l}
x \geqslant-1, \\
\left(\sqrt{x+1}+\frac{1}{2}\right)^{2}>4
\end{array}\right. \\
\Longrightarrow\left\{\begin{array} { l } 
{ x \geqslant - 1 , } \\
{ \sqrt { x + 1 } + \frac { 1 } { 2 } > 2 }
\end{array} \Longrightarrow \left\{\begin{array}{l}
x \geqslant-1, \\
\sqrt{x+1}>\frac{3}{2}
\end{array}\right.\right. \\
\Longrightarrow\left\{\begin{array}{l}
x \geqslant-1, \\
x>\frac{5}{4}
\end{array} \Longrightarrow x>\frac{5}{4}\right. \\
\end{array}
$$

Therefore, the solution to the original inequality is $x>\frac{5}{4}$.