# 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.

• It's strange that you need to prove something about $[0,\frac{16}5]$ when all variables must be equal to $2$. Did you copy down the question correctly? – Théophile Nov 20 '17 at 15:19
• You people are right .This problem may just be wrong. – Abcd Nov 20 '17 at 15:36
• If you need to prove that $e$ is in the mentioned range, simply prove that $e=2$ and that 2 is in the range. – Eric Duminil Nov 20 '17 at 16:43

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 $e$.

• Thanks but can you please explain as to how this inequality is obtained? – Abcd Nov 20 '17 at 15:16
• @Abcd It's a variation on the AM-QM inequality, see the link in my answer. – Arthur Nov 20 '17 at 15:18
• @Abcd It's Cauchy-Schwarz and your given. – Michael Rozenberg Nov 20 '17 at 15:19

\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*}

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 $a = b = c = d = e = 2$.