## The Chebyshev Sum Inequality

If a_{1} ≥ a_{2} ≥ ... ≥ a_{n} and b_{1} ≥ b_{2}
≥ ... ≥ b_{n}
then

n |
_{ n}
**�**
^{
k=1} |
a_{k}b_{k} ≥ |
( |
_{ n}
**�**
^{
k=1} |
a_{k} |
)( |
_{ n}
**�**
^{
k=1} |
b_{k} |
) |

The proof is that the RHS can be written as the sum of n different
sums,

(a_{1}b_{1} + a_{2}b_{2} + ... + a_{n-1}b_{n-1}
+ a_{n}b_{n}) +

(a_{1}b_{2} + a_{2}b_{3} + ... + a_{n-1}b_{n}
+ a_{n}b_{1}) +

...

(a_{1}b_{n} + a_{2}b_{1} + ... + a_{n-1}b_{n-2}
+ a_{n}b_{n-1})

The first of these n sums is at least as big as each of the others by the
rearrangement inequality, and n
times this first sum is the LHS, so the LHS is at least as big as the RHS.

### Related pages in this website

rearrangement inequality

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the Geometric Mean. This is proved using Jensen's Inequality.

The Cauchy-Schwarz inequality

The Triangle Inequality

Puzzles

The Quadratic Formula

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