I wanted to review some fun results in math to practice, so I thought I'd turn it into a fun Christmas blog post. I'm going to talk a little bit about the Cauchy-Schwarz Inequality. It's an exciting result, so stick around.
For any real numbers \((a_1,a_2,...,a_n)\) and \((b_1,b_2,...,b_n)\), the following holds:
$$\left(\sum_{i=1}^{n} a_i b_i\right)^2 \leqslant \left(\sum_{i=1}^{n} a_i^2\right) \left(\sum_{i=1}^{n} b_i^2\right)$$
with equality \(\iff (a_1,a_2,...,a_n)\) and \((b_1,b_2,...,b_n)\) are proportional, i.e., \(\exists\) a constant \(\gamma\) such that \((a_1,a_2,...,a_n)=(\gamma b_1,\gamma b_2,...,\gamma b_n)\).
I'll begin with my favorite (and I think most simple) way to prove the inequality, which leverages properties of quadratics and the discriminant.
Proof 1:
We can consider a quadratic polynomial \(f\) in \(x\):
$$f(x)=\left(\sum_{i=1}^{n} a_i^2\right)x^2-2\left(\sum_{i=1}^{n} a_i b_i\right)x+\left(\sum_{i=1}^{n} b_i^2\right)=\sum_{i=1}^{n} (a_i^2x^2-2xa_ib_i+b_i^2)$$
$$=\sum_{i=1}^{n} (a_ix-b_i)^2$$
Notice now that \(f(x)\geqslant0\) for all \(x\in\mathbb{R}\implies\)the descriminant is negative, i.e.,
$$\left(-2\left[\sum_{i=1}^{n} a_i b_i\right]\right)^2-4\left(\sum_{i=1}^{n} a_i^2\right)\left(\sum_{i=1}^{n} b_i^2\right)=\left(\sum_{i=1}^{n} a_i b_i\right)^2-\left(\sum_{i=1}^{n} a_i^2\right)\left(\sum_{i=1}^{n} b_i^2\right)\leqslant0$$
$$\implies\boxed{\left(\sum_{i=1}^{n} a_i b_i\right)^2\leqslant\left(\sum_{i=1}^{n} a_i^2\right)\left(\sum_{i=1}^{n} b_i^2\right)}$$
Another straightforward proof builds upon the AM-GM equality:
Proof 2:
Let \(A=\sqrt{a_1^2+a_2^2+...+a_n^2}, B=\sqrt{b_1^2+b_2^2+...+b_n^2}\). Then, by applying the AM-GM inequality to the terms \(\frac{a_i^2}{A^2}\) and \(\frac{b_i^2}{B^2}\),
$$\sum_{i=1}^n\frac{a_ib_i}{AB}\leqslant\sum_{i=1}^n\frac{1}{2}\left(\frac{a_i^2}{A^2}+\frac{b_i^2}{B^2}\right)=1$$
which gives
$$\sum_{i=1}^na_ib_i\leqslant AB=\sqrt{a_1^2+a_2^2+...+a_n^2}\sqrt{b_1^2+b_2^2+...+b_n^2}$$
Then, squaring both sides,
$$\boxed{\left(\sum_{i=1}^{n} a_i b_i\right)^2\leqslant\left(\sum_{i=1}^{n} a_i^2\right)\left(\sum_{i=1}^{n} b_i^2\right)}$$
Short, this post is, but I wanted to share something, at least!
-Elijah