I would like to prove Chebyshev's sum inequality, which states that:


If $a\_1\geq a\_2\geq \cdots \geq a\_n$ and $b\_1\geq b\_2\geq \cdots \geq b\_n$, then  

$$
\frac{1}{n}\sum\_{k=1}^n a\_kb\_k\geq \left(\frac{1}{n}\sum\_{k=1}^n a\_k\right)\left(\frac{1}{n}\sum\_{k=1}^n b\_k\right)
$$  

I am familiar with the non-probabilistic proof, but I need a probabilistic one.



How can I prove Chebyshev's sum inequality with probabilistic methods?
Let $A$ be the random variable attaining the values $a\_1,\dotsc,a\_n$ with equal probabilities, and define $B$ similarly, subject to $\mathbb P(B=b\_i|A=a\_i)=1$. Then $\mathbb E(A)=\frac1n\,\sum\_{1\le i\le n} a\_i$, $\mathbb E(B)=\frac1n\,\sum\_{1\le i\le n} b\_i$, and $\mathbb E(AB)=\frac1n\,\sum\_{1\le i\le n} a\_ib\_i$. Since $a\_i$ and $b\_i$ are similarly ordered, we have $\mathrm{Cov}(A,B)>0$ whence $\mathbb E(AB)\ge\mathbb E(A)\mathbb E(B)$.


