Solve the following equation:

$$
\frac{3 x+3}{\sqrt{x}}=4+\frac{x+1}{\sqrt{x^{2}-x+1}} .
$$
Solution. $\sqrt{x}$ appears in the denominator, so $x>0$, furthermore, $x^{2}-x+1$ takes positive values for all real $x$, since

$$
x^{2}-x+1=\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4}
$$

To estimate the left-hand side expression, we use that for $a>0$, $a+\frac{1}{a} \geq 2$. Let's rewrite the left side of the equation and estimate it using the aforementioned inequality:

$$
\frac{3 x+3}{\sqrt{x}}=3\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) \geq 6
$$

Thus, we see that

$$
\begin{array}{r}
4+\frac{x+1}{\sqrt{x^{2}-x+1}} \geq 6 \\
\frac{x+1}{\sqrt{x^{2}-x+1}} \geq 2
\end{array}
$$

We know that $x+1$ is also positive, so we can square the inequality and multiply by the denominator:

$$
\begin{aligned}
x^{2}+2 x+1 & \geq 4 x^{2}-4 x+4 \\
0 & \geq 3 x^{2}-6 x+3 \\
0 & \geq(x-1)^{2}
\end{aligned}
$$

Therefore, the equation can only have $x=1$ as its unique solution. Substitution shows that this is indeed a solution.