Value of $e$ given that $\frac{a+b+c+d+e}{5} = 2 = \sqrt{\frac{a^2 + b^2 + c^2 + d^2 + e^2}{5}}$ I have five real numbers $a,b,c,d,e$ and their arithmetic mean is $2$. I also know that the arithmetic mean of $a^2, b^2,c^2,d^2$, and $e^2$ is $4$. Is there a way by which I can prove that the range of $e$ (or any ONE of the numbers) is $[0,16/5]$. I ran across this problem in a book and  am stuck on it. Any help would be appreciated.

By C-S $$(1^2+1^2+1^2+1^2)(a^2+b^2+c^2+d^2)\geq(a+b+c+d)^2$$ or
$$4(a^2+b^2+c^2+d^2)\geq(a+b+c+d)^2$$ or
$$4(20-e^2)\geq(10-e)^2$$ or
$$(e-2)^2\leq0$$ or
$$e=2.$$
This method works in the general case.
Given: $a+b+c+d+e=k$ and $a^2+b^2+c^2+d^2+e^2=l$.
Find the range of $e$. 
