$a,b,c,d,e$ are positive real numbers such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$, find the range of $e$. 
$a,b,c,d,e$ are positive real numbers such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$, find the range of $e$.

My book tells me to use tchebycheff's inequality
$$\left(\frac{a+b+c+d}{4}\right)^2\le \frac{a^2+b^2+c^2+d^2}{4}$$
But this not the Chebyshev's inequality given in wikipedia. Can someone state the actual name of the inequality so I can read more about it?
(I got $e\in\left[0,\frac{16}{5}\right]$ using the inequality)

As @ChenJiang stated, its a case of cauchy's inequality
$$\left(\frac{a+b+c+d}{4}\right)^2\le \frac{a^2+b^2+c^2+d^2}{4}$$
$$(a+b+c+d)^2\le 4(a^2+b^2+c^2+d^2)$$
$$(8-e)^2\le  4(16-e^2)$$
$$5e^2-16e\le 0$$
$$e(5e-16)\le 0$$
$$\implies 0\le e\le \frac{16}{5}$$
