Let $y = nf(x)$ for some integers $n$ and $x$
Let $z = f(a_1x)+f(a_2x)+f(a_3x) +...+ f(a_nx)$, with $a_1+a_2 +..+ a_n = n$, and for some integer $i$ ($0<i<n+1$), $a_i\neq1$.
Prove that $y>z$.
I feel like this shouldn't be too difficult, but I'm not sure how to proceed.
Or, equivalently (I believe, correct me if I'm wrong), prove that $U = f(d_1) + f(d_2) +...+ f(d_n)$, with $d_1 + d_2 +...+ d_n = x$, is maximized when $d_1 = d_2 =...= d_n = x/n$.
I think I can prove this version using a Lagrangian multiplier, but in that proof, I'm not sure where I use the fact that $f$ is concave (as the claim is certainly false if it's convex). Anyway, I'm more interested in a proof using the latter point of view. Thanks!
Here's the main idea. A convex function $g$ is one where for all $x,y$ in the domain, $$ g(tx+(1-t)y)\leq tg(x)+(1-t)g(y) $$ holds for all $0\leq t \leq 1$. By this definition, you can prove the following by induction. If $\lambda_1 + \lambda_2+...+\lambda_n=1$, then for $n$ points $x_1,..,x_n$, $$ g(\lambda_1x_1+...+\lambda_nx_n)\leq \lambda_1g(x_1)+...+\lambda_ng(x_n). $$ Interesting, this is actually Jensen's inequality for a discrete random variable with $n$ possible outcomes (sum of probabilities is 1).
In your case, proving $y>z$ with $f$ concave is the same as proving $-f(x)<-z/n$ with $-f$ convex. Just set $g=-f$. Then by the argument above involving convexity, it is clear that $\lambda_i=1/n$ and $x_i=a_ix$.