# Discrete Chebyshev inequality as double projection

Let $$a_1< a_2<\ldots< a_n$$ and $$b_1, b_2, \ldots, b_n$$ be real numbers. Then \begin{align*} &b_1< b_2< \ldots< b_n\Rightarrow \frac{a_1b_1 + a_2b_2 +\ldots + a_nb_n}{n}> \frac{a_1 + a_2 + \ldots + a_n}{n}\cdot\frac{b_1 + b_2 + \ldots + b_n}{n},\\ &b_1> b_2> \ldots> b_n\Rightarrow \frac{a_1b_1 + a_2b_2 +\ldots + a_nb_n}{n} < \frac{a_1 + a_2 + \ldots + a_n}{n}\cdot\frac{b_1 + b_2 + \ldots + b_n}{n}. \end{align*} Another way to view Chebyshev's Inequality: $$\frac{\vec a\cdot\vec b}{n}\geq \frac{\vec a\cdot\vec u}{n}\cdot\frac{\vec b\cdot\vec u}{n},$$ where $$\vec u = (1,1,\ldots,1)$$. If $$\theta$$ is the angle between $$\vec a$$ and $$\vec b$$ and $$\alpha$$ and $$\beta$$ are the angles between $$\vec a$$ and $$\vec u$$ and $$\vec b$$ and $$\vec u$$, respectively, we have $$|\vec a|\cdot|\vec b|\cos\theta\geq |\vec a|\cdot|\vec b|\cos\alpha\cos\beta\Leftrightarrow\cos\theta\geq\cos\alpha\cos\beta.$$ The LHS represents projection of a unit vector onto some line $$l$$, while RHS represents projection of a unit vector onto some third line $$m$$ and then projection of that image onto $$l$$. I'm looking for a neat intuitive explanation of this inequality, but I'll settle for explanations that involve higher math. Also, there should be a generalization: Let $$X,Y,Z$$ be spaces and let $$s$$ be a set in $$X$$. Then projection of $$s$$ onto $$Z$$, $$\pi_Z(s)$$, has volume at least as large as $$\pi_Z(\pi_Y(s))$$. Thanks.

## 1 Answer

I have an answer to the special case: First, assume that $$\alpha + \beta = \theta$$. I.e., that the intermediate projection line $$m$$ lies in the same plane as the unit vector and $$l$$. Then $$\cos\theta = \cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta \geq\cos\alpha\cos\beta,$$ since $$\alpha,\beta\in[0,\pi]$$ and $$\sin x$$ is non-negative in that interval. Now as we continuously move $$m$$ out of the plane, $$\alpha + \beta$$ increases to at most $$2\pi - \theta$$ and $$\cos x \leq\cos\theta$$ for $$x\in[\theta,2\pi-\theta]$$. However, it would be nice to have a more intuitive solution as well as a generalization.