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> Let $ 0<a<b $, $ f\in C^1\left([0,b]\right)$. Assume that $ f $ is concave on $ [0,a] $ and convex on $ [a,b] $ with $ f'(0)>f'(b) $. Please prove that there exist $ n\_0\in\mathbb{N} $ which is sufficiently large, such that for any $ n\geq n\_0 $ and $ a\_1+a\_2+...+a\_n=b $ with $ a\_i\geq 0 $ ($ i=1,2,...,n $), we have $ \sum\_{i=1}^nf(a\_i)\leq nf\left(\frac{b}{n}\right) $.
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Simple observation shows that if $ a\_i\in [0,a] $ for all $ i $, the result is easy to prove. However, I cannot deal with the situation that there exists some $ i\_0 $ such that $ a\_{i\_0}\in(0,b] $. I guess that such case is related with the statement that $ f'(0)>f'(b) $, but I do not know how to continue the proof. Could you give me some references or hints?



How to prove $ \sum_{k=1}^{n}f(a_k)\leq nf\left(\frac{b}{n}\right) $ for sufficiently large $ n $ here?
Denote $b/n=s$. We want a pointwise bound $$f(x)\leqslant f(s)+(x-s)f'(s),\label{1}\tag{$\heartsuit$}$$ then summing \eqref{1} up for $x=a\_1,\ldots,a\_n$ we get the desired inequality. Note that if $s<a$ (that holds for $n>b/a$) we get \eqref{1} on $[0,a]$ by concavity. For proving \eqref{1} on $[a,b]$, by convexity it suffices to verify \eqref{1} for $x=a$ and $x=b$. For $x=a$ this is already done, for $x=b$ it reads as
$$
f(b)\leqslant f(s)+(b-s)f'(s).
$$ When $n$ is large, RHS converges to $f(0)+bf'(0)$. Thus it suffices to check that
$$
f(b)<f(0)+bf'(0).
$$ Assume the contrary:
$$
f(b)\geqslant f(0)+bf'(0).
$$ We have $f(x)\leqslant f(0)+f'(0)x$ for all $x\in [0,a]$ by concavity. Denote by $c$ the endpoint of the maximal segment $[0,c]$ on which we have $f(x)\leqslant f(0)+f'(0)x$. Then $c\in [a,b]$ and we have $f(c)=f(0)+f'(0)c$ (otherwise $c$ is not maximal). This yields
$$
f'(c)=\lim\_{x\to c-0}\frac{f(c)-f(x)}{c-x}\geqslant f'(0).
$$ Since $f'$ increases on $[a,b]$ by convexity we get $f'(b)\geqslant f'(c)\geqslant f'(0)$, a contradiction.


