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The Cauchy-Schwarz inequality states $|\langle x,y \rangle |\leq ||x||\cdot ||y||.$ Use this tag for questions related to the CS inequality and its applications.

13 votes

Is there a combinatorial proof of Cauchy-Schwarz?

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 …
Igor Pak's user avatar
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