8. let \(a, b, c, d\) and \(e\) be positive real numbers. Determine the largest value that the following expression can take:

\[
\frac{a b + b c + c d + d e}{2 a^{2} + b^{2} + 2 c^{2} + d^{2} + 2 e^{2}}
\]
Solution (Arnaud): The idea is clearly to apply AM-GM. We must therefore decompose $b^{2}=$ $x b^{2}+y b^{2}$ and probably $2 c^{2}=c^{2}+c^{2}$. The numbers $x, y$ must therefore satisfy $x+y=1$ and $2 x=y$. We get $x=1 / 3$ and $y=2 / 3$. By AM-GM, we obtain

$$
\begin{aligned}
2 a^{2}+1 / 3 b^{2} & \geq 2 \sqrt{2 / 3} \cdot a b \\
2 / 3 b^{2}+c^{2} & \geq 2 \sqrt{2 / 3} \cdot b c \\
c^{2}+2 / 3 d^{2} & \geq 2 \sqrt{2 / 3} \cdot c d \\
1 / 3 d^{2}+2 e^{2} & \geq 2 \sqrt{2 / 3} \cdot e d
\end{aligned}
$$

The maximum is therefore $(2 \sqrt{2 / 3})^{-1}=\sqrt{3 / 8}$. The equality cases are given by the applications of AM-GM. We get, for example, $a=1, b=\sqrt{6}, c=2, d=\sqrt{6}$ and $e=1$.