Usbrth

Let $a_1< a_2<ldots< a_n$ and $b_1, b_2, ldots, b_n$ be real numbers. Then
beginalign*
&b_1< b_2< ldots< b_nRightarrow fraca_1b_1 + a_2b_2 +ldots + a_nb_nn> fraca_1 + a_2 + ldots + a_nncdotfracb_1 + b_2 + ldots + b_nn,\
&b_1> b_2> ldots> b_nRightarrow fraca_1b_1 + a_2b_2 +ldots + a_nb_nn < fraca_1 + a_2 + ldots + a_nncdotfracb_1 + b_2 + ldots + b_nn.
endalign*
Another way to view Chebyshev's Inequality:
$$
fracvec acdotvec bngeq fracvec acdotvec uncdotfracvec bcdotvec un,
$$
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
$$
|vec a|cdot|vec b|costhetageq |vec a|cdot|vec b|cosalphacosbetaLeftrightarrowcosthetageqcosalphacosbeta.
$$
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.

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
$$
costheta = cos(alpha+beta) = cosalphacosbeta - sinalphasinbeta geqcosalphacosbeta,
$$
since $alpha,betain[0,pi]$ 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 $2pi - theta$ and $cos x leqcostheta$ for $xin[theta,2pi-theta]$.
However, it would be nice to have a more intuitive solution as well as a generalization.