Cauchy’s inequality and the appeal to symmetry

Working over the reals $\mathbb{R}$ a young mathematician will learn the inequality $(a_1b_1 + \cdots + a_nb_n)^2 \leq (a_1^2 + \cdots + a_n^2)(b_1^2 + \cdots + b_n^2)$ which has its name associated to Cauchy and Schwarz and is called the Cauchy -Schwarz inequality.

After proving this result the question of when this inequality is an equality is brought up and the young mathematician will learn this occus when $a_i = \lambda b_i$ for some $\lambda \in \mathbb{R}$ and for all $i \in \{1,\cdots,n \}$ . Here is a neat niave proof of this result following the mathematicians’ philosophy of ‘look for symmetry’.

Let us look at the ‘error’ $E_n = (a_1^2 + \cdots + a_n^2)(b_1^2 + \cdots + b_n^2) - (a_1b_1 + \cdots + a_nb_n)^2.$

Expanding out gives us $E_n = \sum_{}^{}a_i^2b_j^2 - \sum_{}^{}a_ib_ia_jb_j$

And since $\sum_{}^{}a_i^2b_j^2 = \frac{1}{2}\sum_{}^{}(a_i^2b_j^2 + a_j^2b_j^2).$ Yes I did just write this and no, I’m not insulting your inteligence. This is my apeal to symmetry.

We get $E_n = \frac{1}{2}\sum_{}^{}(a_i^2b_j^2 + a_j^2b_j^2) - \sum_{}^{}a_ib_ia_jb_j = \frac{1}{2}\sum_{}^{}(a_i^2b_j^2 + a_j^2b_j^2 - 2a_ib_ia_jb_j )$

And so $E_n = \frac{1}{2}\sum_{}^{}(a_ib_j - a_jb_i)^2.$ I’ll let you finish the rest off but we’ve done all the hard work here.

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