$C_n = \frac{1}{2n}$. Example for proving maximal $a_1=2; a_2=1; a_3=1; ....; a_{n-1}=1; a_n=0$. Let's prove that for such $C_n$ inequality holds. $\frac{a_1^2+...+a_n^2}{n}-( \frac{a_1+...a_n}{n})^2-\frac{1}{2n} (a_1-a_n)^2$ = $\frac{(\frac{a_1+a_n}{2})^2+a_2+a_3+....+a_{n-1}+(\frac{a_1+a_n}{2})^2}{n}-$ $(\frac{\frac{a_1+a_n}{2}+a_2+a_3+....+a_{n-1}+\frac{a_1+a_n}}{2})^2 \geq 0 $ by Cauchy for numbers $(\frac{a_1+a_n}{2};a_2;a_3;....;a_{n-1};\frac{a_1+a_n}{2})$ So maximal constant is $\frac{1}{2n}$ By generalization of this problem and other generalizations of Power mean inequalities project on Intel ISEF 2010 "Generalization of Power mean inequality"

Martin N. wrote: For each integer $n\geqslant2$, determine the largest real constant $C_n$ such that for all positive real numbers $a_1, \ldots, a_n$ we have \[\frac{a_1^2+\ldots+a_n^2}{n}\geqslant\left(\frac{a_1+\ldots+a_n}{n}\right)^2+C_n\cdot(a_1-a_n)^2\mbox{.}\] (4th Middle European Mathematical Olympiad, Team Competition, Problem 2) if $a_1=a_n$ then $C_n$ can be any reals. Let us assume $a_1\ne a_n$, then we just need to find the min of $f\left( {a_1 , \ldots ,a_n } \right) = \frac{{\frac{{a_1^2 + \ldots + a_n^2 }}{n} - \left( {\frac{{a_1 + \ldots + a_n }}{n}} \right)^2 }}{{\left( {a_1 - a_n } \right)^2 }}$ let $a_1=a,a_2+...+a_{n-1}=(n-2)b,a_n=c$, then $a\ne c$ and by Cauchy Inequality, we have $f\left( {a_1 , \ldots ,a_n } \right) \ge \frac{{\frac{{a^2 + \left( {n - 2} \right)b^2 + c^2 }}{n} - \left( {\frac{{a + \left( {n - 2} \right)b + c}}{n}} \right)^2 }}{{\left( {a - c} \right)^2 }}$ $ = \frac{{\left( {n - 2} \right)\left( {a + c - 2b} \right)^2 + n\left( {a - c} \right)^2 }}{{2n^2 \left( {a - c} \right)^2 }} \ge \frac{1}{{2n}}$ when $a_2 = a_3 = \cdots = a_{n - 1} = \frac{{a_1 + a_n }}{2}$ equality holds, so $C_n=\frac{1}{2n}$.

this is China shanxi olipic 设a1,a2,…,an是正数，min{a1,a2,…,an}＝a1，max{a1,a2,…,an}＝an，证明不等式： a1^2＋a2^2＋…＋an^2≥1/n(a1＋a2＋…＋an)^2＋1/2(a1－an)^2.

kuing wrote: Martin N. wrote: For each integer $n\geqslant2$, determine the largest real constant $C_n$ such that for all positive real numbers $a_1, \ldots, a_n$ we have \[\frac{a_1^2+\ldots+a_n^2}{n}\geqslant\left(\frac{a_1+\ldots+a_n}{n}\right)^2+C_n\cdot(a_1-a_n)^2\mbox{.}\] (4th Middle European Mathematical Olympiad, Team Competition, Problem 2) if $a_1=a_n$ then $C_n$ can be any reals. Let us assume $a_1\ne a_n$, then we just need to find the min of $f\left( {a_1 , \ldots ,a_n } \right) = \frac{{\frac{{a_1^2 + \ldots + a_n^2 }}{n} - \left( {\frac{{a_1 + \ldots + a_n }}{n}} \right)^2 }}{{\left( {a_1 - a_n } \right)^2 }}$ let $a_1=a,a_2+...+a_{n-1}=(n-2)b,a_n=c$, then $a\ne c$ and by Cauchy Inequality, we have $f\left( {a_1 , \ldots ,a_n } \right) \ge \frac{{\frac{{a^2 + \left( {n - 2} \right)b^2 + c^2 }}{n} - \left( {\frac{{a + \left( {n - 2} \right)b + c}}{n}} \right)^2 }}{{\left( {a - c} \right)^2 }}$ $ = \frac{{\left( {n - 2} \right)\left( {a + c - 2b} \right)^2 + n\left( {a - c} \right)^2 }}{{2n^2 \left( {a - c} \right)^2 }} \ge \frac{1}{{2n}}$ when $a_2 = a_3 = \cdots = a_{n - 1} = \frac{{a_1 + a_n }}{2}$ equality holds, so $C_n=\frac{1}{2n}$. $\frac{{\left({n-2}\right)\left({a+c-2b}\right)^{2}+n\left({a-c}\right)^{2}}}{{2n^{2}\left({a-c}\right)^{2}}}\ge\frac{1}{{2n}} $ How you get this?

$ \frac{{\left({n-2}\right)\left({a+c-2b}\right)^{2}+n\left({a-c}\right)^{2}}}{{2n^{2}\left({a-c}\right)^{2}}}\ge{{\left0+n\left({a-c}\right)^{2}}}{{2n^{2}\left({a-c}\right)^{2}}}

I don't like previous solutions because Ovchinnikov Denis didn't prove that $C_n=\frac{1}{2n}$ is maximal constant (or at least I can't see how), and kuing probably guessed $C_n$ and then proved it. I have systematical way to prove that constant. WLOG we can suppose that $a_2,...,a_{n-1} \in [a_1,a_n]$ and since inequality is symmetric in variables $a_2,...,a_{n-1}$, we can also suppose $a_2\le ... \le a_{n-1}$. Let's make a change $(a_2,a_{n-1}) \rightarrow (\frac{a_2+a_{n-1}}{2},\frac{a_2+a_{n-1}}{2})$. With this change, RHS stays untouched, so we need to prove: $\frac{a_1^2+a_2^2+...+a_{n-1}^2+a_n^2}{n} \ge \frac{a_1^2+(\frac{a_2+a_{n-1}}{2})^2+...+(\frac{a_2+a_{n-1}}{2})^2+a_n^2}{n}$ what is clearly true by Cauchy-Schwarz inequality. Let $a_1=1$ (because of homogenity), and $a_n=a$. Using mixing variable method, we just need to prove inequality for case $t=a_2=...=a_n=\frac{a_2+...+a_{n-1}}{n-2}$, $t\in [1,a]$. Inequality is equal to: $\frac{a^2+(n-2)t^2+1}{n} \ge \left( \frac{1+(n-2)t+a}{n} \right)^2+ C_n(a-1)^2$ After expanding we have: $2(n-2)(t-\frac{a+1}{2})^2+(K_1\cdot a^2+K_2\cdot a +K_3) \ge 0$ Where $K_1=-\frac{n-2}{2}+n-1-n^2C_n$. We see that $K_1$ must be greater than zero

(see why)

and that means $-\frac{n-2}{2}+n-1-n^2C_n \ge 0$ $\Leftrightarrow$ $C_n \le \frac{1}{2n}$. Now we just need to prove inequality for $C=\frac{1}{2n}$ what Ovchinnikov Denis has done very well. EDIT to fix a typo.

Ineqality holds for all numbers => it holds for numbers $a_1=3; a_2=2; a_3=2; .... a_{n-1}=2; a_n=1 $. Then inequality becomes $\frac{9+4(n-1)+1}{n} \geq (\frac{3+2n+1}{n})^2 + C_n(3-1)^2 <=> 1+ \frac{2}{n} \geq 1+ C_n 4 <=> C_n \leq \frac{1}{2n}$ (In my first post I write, that equality holds for $a_1=2; a_2=1; a_3=1; .....; a_{n-1}=1; a_n=0$, but $a_n=0$ in this example, so it is not true ) But this example is correct !!

Now it is true, but that is still guessing the constant.

We write the inequality as $ \frac{\sum a ^2}{n} - (\frac{\sum a}{n})^2\geq C_n(a_n -a_1)^2 $ .The left hand side can be written as $ \frac{\sum (a-\alpha)^2 }{n} $ where $ \alpha $ is the arithmetic mean of $ a_1,...,a_n $.We substitute $ x_i=a_i-\alpha $.Then the inequality is equivalent to $ \frac{\sum x^2 }{n}\geq C_n(x_n -x_1)^2 $,and we have $ \sum x=0 $.By setting $ x_1=-c,x_2=...=x_{n-1}=0,x_n=c $ we find that $ C_n\leq \frac{1}{2n} $.Thus we have to prove that $ \frac{\sum x^2 }{n}\geq \frac{1}{2n}(x_n -x_1)^2 $ .It is clear that $ x_i^2\geq 0 $ so it remains to show that $ \frac{x_1^2+x_n^2}{n}\geq \frac{1}{2n}(x_n -x_1)^2 $ which is clearly equivalent to $ (x_1+x_n)^2\geq 0 $.QED

The inequality is equivalent to: \[((n-2)(a_1+a_n)-2(a_2+\ldots+a_{n-1}))^2+2n((n-2)(a_2^2+\ldots+a_{n-1}^2)-(a_2+\ldots+a_{n-1})^2) \ge n(n-2)(2nC_n-1)(a_1-a_n)^2\] Clearly, the $LHS$ is always nonnegative. Setting $C_n=\frac{1}{2n}$ makes the $RHS$ zero, thus satisfying the inequality. Equality holds for $a_1+a_n=2a_2=\ldots=2a_{n-1}$,

Solved with Alfreed . Set $a_1=\frac{1}{2}, a_2=a_3=\ldots=a_{n-1}=1, a_{n}=\frac{3}{2}$ to get that $C_n \le \frac{1}{2n}$. Now we will prove that: \[\frac{a_1^2+\ldots+a_n^2}{n}\geqslant\left(\frac{a_1+\ldots+a_n}{n}\right)^2+\frac{1}{2n}\cdot(a_1-a_n)^2\mbox{.}\]Since inequality is homogenous we can assume that $a_1+a_2+\ldots+a_n=n$. We can rewrite inequality as follows: \begin{align*} 2(a_1^2+a_2^2+\ldots+a_n^2) \ge 2n + (a_1-a_n)^2 \\ 2(a_1^2+a_n^2)+2(a_2^2+\ldots+a_{n-1}^2) \ge 2n + (a_1-a_n)^2 \end{align*}From Cauchy inequality we have: $$ (a^2+\ldots+a_{n-1}^2) \ge \frac{(a_2+\ldots+a_{n-1})^2}{n-2} $$Therefore: $$ 2(a_1^2+a_n^2)+2(a_2^2+\ldots+a_{n-1}^2) \ge 2(a_1^2+a_n^2)+2\frac{(a_2+\ldots+a_{n-1})^2}{n-2} $$It remains to show that: \begin{align*} 2(a_1^2+a_n^2)+2\frac{(a_2+\ldots+a_{n-1})^2}{n-2} \ge 2n + (a_1-a_n)^2 \\ (a_1+a_n)^2 + 2\frac{(a_2+\ldots+a_{n-1})^2}{n-2} \ge 2n \\ (a_1+a_n)^2+2\frac{(n-(a_1+a_n))^2}{n-2} \ge 2n \\ x^2(n-2)+2(n-x)^2 \ge 2n(n-2) \\ x^2n-2x^2+2n^2-4nx+2x^2 \ge 2n^2-4n \\ x^2n -4nx+4n \ge 0 \\ n(x-2)^2 \ge 0 \end{align*}where $x=a_1+a_n$. Hence the conclusion.