I do not know whether this counts, but you may do the following.

At first, you may suppose that $a_n>0$ (otherwise replace all $a_i$ to $a_i-a_n+1$). Consider the probabilistic distribution $\mu$ on $\{1,\ldots,n\}$ such that $p_i={\rm prob} (X=i)=a_i/(a_1+\ldots+a_n)$. Then $p_1+\ldots+p_n=1$, $p_1\geqslant \ldots \geqslant p_n$, and we want to prove that $\mathbb{E}_{\mu} b\geqslant \mathbb{E}_{\lambda} b\,\, (*),$ where $\lambda$ is a uniform distribution, and $b:i\to b_i$ is a decreasing function. For doing this we construct a coupling, that is, a joint distribution $\nu$ of $(X,Y)\in\{1,\ldots,n\}^2$ such that $X$ is distributed as $\mu$, $Y$ as $\lambda$ and $Y\geqslant X$. This gives $(*)$ since $$\mathbb{E}_{\mu} b=\mathbb{E}_\nu b(X)\geqslant \mathbb{E}_\nu b(Y)=\mathbb{E}_\lambda b.$$
For constructing $\nu$, partition the semiinterval $[0,1)$ onto $n$ semiintervals $\Delta_1,\ldots,\Delta_n$ (from left to right) of lengths $1/n$ and semiintervals $\delta_1,\ldots,\delta_n$ of lengths $p_1,\ldots,p_n$. Choose a random point $x\in [0,1]$ at uniform and denote $(X,Y)=(k,i)$ if $x\in \Delta_i\cap \delta_k$.

It remains to prove that $i\geqslant k$. Assume the contrary: $k>i$. Then $x\geqslant p_1+\ldots+p_i\geqslant ip_i$; $1-x\leqslant p_{i+1}+\ldots+p_n\leqslant (n-i)p_i$; $x/i\geqslant p_i\geqslant (1-x)/(n-i)$; $x\geqslant i/n$; $x\notin \Delta_i$. A contradiction.

physicsproof (relying on the mechanics of the center of mass), see Picard’s Weighty Proof of Chebyshev’s Sum Inequality $\endgroup$ – Carlo Beenakker Jan 13 at 10:22