it is easy problem by http://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=113248&p=643414&hilit=smv#p643414

By Karamata inequality this reduces to proving that if $a_1+...+a_{n+1}=0$ and we let $a$ be that sequence, and $-a/n$ denote the sequence $(-a_1/n)+...+(-a_{n+1}/n)=0$, we must prove $a$ majorizes $-a/n$ WLOG $a_1 \ge ... \ge a_n$. Let $X_i=a_i+...+a_{i+x-1}$ for a given $1 \le x \le n+1$, with indices mod $n+1$. Notice that, if $b_i=-a{n+2-i}/n$ then $b_1 \le ... \le b_n$ and $0 = X_1+...+X_n \le nX_1+X_{n+2-x} \Rightarrow a_1+...+a_x = X_1 \ge X_{n+2-x}/n = b_1+...+b_x$ for all $x$, so we're done.

Is thwre solution by AM GM??

ThisIsART wrote: Is thwre solution by AM GM?? see my post .. my solution is just AM-GM

I do not understand i am sorry give complete solution please

Let $f(x)=n^{e^x}$. Since $f$ is a convex function, we can use Karamata.

I have not study about convex and concaf function i am sorry sir

Aryan-23 wrote:

Nice Solution.