Prove that $\frac{8}{5}\le 2a+b\le 8$ Let $a,b,c,d,e$ be real numbers such that
$$\begin{cases}
a+b+c+d+e=8\\
a^2+b^2+c^2+d^2+e^2=16
\end{cases}$$
Prove that:
$$\dfrac{8}{5}\le 2a+b\le 8$$

Use Cauchy-Schwarz inequality we have
$$3(c^2+d^2+e^2)\ge (c+d+e)^2\Longrightarrow 48-3a^2-3b^2\ge (8-a-b)^2$$
then let $2a+b=t$, you have
$$8a^2+(8-7t)a+2t^2-8t+8\le 0$$
$$\Longrightarrow \Delta_{a}=(8-7t)^2-32(2t^2-8t+8)\ge 0\Longrightarrow \dfrac{8}{5}\le t\le 8$$
