# Is there a combinatorial proof of Cauchy-Schwarz?

I've only played with this a little for the past day or so, and haven't thought about it too hard, so it might be obvious. Obviously it's not fair to ask for a "combinatorial proof" of an inequality involving real numbers, so we'll ask that the vectors be in $\mathbb{N}^n$. More concretely:

Given n boxes subdivided into a "right half" and a "left half" with $a_i$ objects in the right half of box $i$, and $b_i$ in the left half of box i, is there a natural injective function from

Two pairs (ordered, with replacement) of objects, with each pair containing one object from the left half and one object from the right half of a fixed box

to

A pair (ordered, with replacement) from the right half of some box, and a pair (O,WR) from the left half of some (possibly different) box?

(Sorry if this is a double; my wireless is being strange.)

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Cauchy-Schwarz is true for reals if and only if it's true for rationals by continuity, if and only it's true for integers by homogeneity. – Qiaochu Yuan Feb 23 '10 at 7:02
Also, have you seen the discussion at golem.ph.utexas.edu/category/2007/04/… ? – Qiaochu Yuan Feb 23 '10 at 7:08
The discussion continued at golem.ph.utexas.edu/category/2007/05/…. I think the conclusion was that categorification of Cauchy-Schwarz was not possible. – David Corfield Feb 23 '10 at 9:02
I haven't finished reading it, but if you're interested in anything and everything Cauchy-Schwarz related, pick up The Cauchy-Schwarz Master Class by J. Michael Steele. – Ian Durham Feb 23 '10 at 12:30

Honestly, not really, there is no interesting combinatorial proof. Think of the most trivial case $a^2 + b^2 \ge 2 a b$. How do you prove that combinatorially? Well, arrange the terms into $(a+b)^2$ as in proving Pythagoras theorem and cut both squares by a diagonal. Now observe that reflection of white triangles will cover the blue rectangles. This proof can be stated in a purely combinatorial language, but why bother - it's a standard "book proof" you can find everywhere. Similarly, you can take: $$\sum_{i=1}^n \sum_{j=1}^n \left(a_i b_j + a_j b_i \right)^2$$ expand all the terms and get a similar flavor "combinatorial proof", more or less the same what you can find in standard books. But again why bother?