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.
 A: If we have
$$
\frac{a+b+c+d+e}{5} = 2 = \sqrt{\frac{a^2 + b^2 + c^2 + d^2 + e^2}{5}}
$$
then all the five numbers are necessarily equal to $2$, as dictated by the AM-QM inequality.
PS. Technically, the power mean inequality is only valid for positive real numbers, but if any of the numbers were negative, then we could change its sign and increase the arithmetic mean without changing the quadratic mean, and the quadratic mean would still be larger. So if the two means are equal, even if allowing for negative numbers, we still get $a = b = c = d = e = 2$.
A: 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 $e$. 
A: \begin{align*}
&(a - 2)^2 + (b-2)^2 + (c-2)^2 + (d-2)^2 + (e - 2)^2\\[4pt]
&= (a^2 + b^2 + c^2 + d^2 +e^2) - 4(a + b + c + d +e) + 20\\[4pt]
&= 20 - 4(10) + 20\\[4pt]
&= 0\\[10pt]
&\;\text{hence}\\[10pt]
&\;a = b = c = d = e = 2\\[4pt]
\end{align*}
