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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.

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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.

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