<s>[INST] 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.
 [/INST] \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*}
</s>