8. Problem: Find all real numbers $x$ such that $-1<x \leq 2$ and
$$
\sqrt{2-x}+\sqrt{2+2 x}=\sqrt{\frac{x^{4}+1}{x^{2}+1}}+\frac{x+3}{x+1} .
$$
Solution: (Viet Hoang)
Notice that we are solving for $x$ in the domain $-1<x \leq 2$. Using Cauchy-Schwarz Inequality on the left hand side, one has
$$
\sqrt{2-x}+\sqrt{2+2 x}=\sqrt{\frac{1}{2} \cdot(4-2 x)}+\sqrt{2+2 x} \leq \sqrt{\left(\frac{1}{2}+1\right)(2+2 x+4-2 x)}=3 .
$$

Using the fact that $x^{4}+1 \geq 2 x^{2}$ (because $\left(x^{2}-1\right)^{2} \geq 0$ ) and $x^{2}+1 \geq 2 x$ (because $(x-1)^{2} \geq 0$ we can get:
$$
2\left(x^{4}+1\right) \geq\left(x^{2}+1\right)^{2} \geq\left(x^{2}+1\right) \frac{(x+1)^{2}}{2} .
$$

If we substitute this into the right-hand side we get
$$
\sqrt{\frac{x^{4}+1}{x^{2}+1}}+\frac{x+3}{x+1} \geq \frac{x+1}{2}+\frac{x+3}{x+1}=\frac{x+1}{2}+\frac{2}{x+1}+1 .
$$

Finally applying the AM-GM again for 2 positive numbers $\frac{x+1}{2}$ and $\frac{2}{x+1}$ we obtain $\frac{x+1}{2}+$ $\frac{2}{x+1} \geq 2$. Thus the right-hand side is $\geq 3$.

Therefore, we can see that equality must occur in our AM-GM inequality and hence, $\frac{x+1}{2}=\frac{2}{x+1} \Longrightarrow x=1$