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:
$$\dfrac{8}{5}\le 2a+b\le 8$$
 A: 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
$$8a^2+(8-7t)a+2t^2-8t+8\le 0$$
$$\Longrightarrow \Delta_{a}=(8-7t)^2-32(2t^2-8t+8)\ge 0\Longrightarrow \dfrac{8}{5}\le t\le 8$$
A: Let we set $(a,b,c,d,e)=\frac{8}{5}\cdot(1,1,1,1,1)+(x_1,x_2,x_3,x_4,x_5)$.
Then we have:
$$\left\{\begin{array}{rcl}x_1+x_2+x_3+x_4+x_5&=&0,\\ x_1^2+x_2^2+x_3^2+x_4^2+x_5^2&=&\frac{16}{5}\end{array}\right.$$
and we have to find the stationary points of $2x_1+x_2$. Lagrange multipliers give:
$$ (2,1,0,0,0) = \lambda(1,1,1,1,1)+2\mu(x_1,x_2,x_3,x_4,x_5)$$
so in the stationary points we have $x_3=x_4=x_5=t$, $\lambda+2\mu t=0$,
$$ 2\mu x_1 = 2+2\mu t,\qquad 2\mu x_2 = 1+2\mu t $$
then:
$$ (x_1,x_2,x_3,x_4,x_5) = \left(\frac{1}{\mu}+t,\frac{1}{2\mu}+t,t,t,t\right) $$
or:
$$ (x_1,x_2,x_3,x_4,x_5) = \left(-\frac{7}{3},-\frac{2}{3},1,1,1\right)\cdot t = \left(-\frac{7}{3},-\frac{2}{3},1,1,1\right)\cdot \frac{\pm 3}{5}$$
so that:
$$ -\frac{16}{5}\leq 2x_1+x_2 \leq \frac{16}{5}$$
and:

$$ \frac{8}{5}\leq 2a+b \leq 8 $$

as wanted, with equality attained by $\left(a,b,c,d,e\right) = \frac{8}{5}(1,1,1,1,1)\pm\frac{1}{5}\left(-7,-2,3,3,3\right).$
