Solve for real $x$ :

$$
x+\sqrt{(x+1)(x+2)}+\sqrt{(x+2)(x+3)}+\sqrt{(x+3)(x+1)}=4 .
$$
First, note by monotonicity of LHS there is exactly one solution.

Let $z=x+2$ and add 2 to both sides. Then we obtain

$$
(\sqrt{z}+\sqrt{z+1})(\sqrt{z}+\sqrt{z-1})=6 .
$$

Then, we can multiply through to get

$$
\begin{aligned}
\sqrt{z}+\sqrt{z-1} & =6(\sqrt{z+1}-\sqrt{z}) \\
7 \sqrt{z} & =6 \sqrt{z+1}-\sqrt{z-1} \\
49 z & =36(z+1)+(z-1)-12 \sqrt{z^{2}-1} \\
12 z & =35-12 \sqrt{z^{2}-1} \\
(12 z-35)^{2} & =12^{2}\left(z^{2}-1\right) \\
35^{2}+12^{2} & =2 \cdot 12 \cdot 35 z .
\end{aligned}
$$

Solving gives $z=\frac{1369}{840}$, hence $x=-\frac{311}{840}$.