How can I prove the Cauchy-Schwarz inequality using Lagrange Multipliers

Question

I do not understand this test, could you do it with more details?

Answer 1

I think the wording of the proof you have attached above is what you are having difficulty with. Let me show a much simpler worded translation of the same proof below, which is also a well known application of Lagrange multipliers.

Let $\mathbf{a} = (a_1,.....,a_n) \in \mathbb{R}^n,$ and $\mathbf{b} = (b_1,....., b_n) \in \mathbb{R}^n.$

We need to show that $\sum_{i=1}^{n} a_i b_i \le || \mathbf{a} || \cdot || \mathbf{b} ||.$

We can homogenize and assume without losing any generality that $|| \mathbf{a} || = 1.$

So the problem reduces to that of showing for any fixed $\mathbf{b} \in \mathbb{R}^n,$ we have $|| \mathbf{a} || = 1 \implies \sum_{i=1}^{n} a_i b_i \le || \mathbf{b} ||.$

Let us put this problem in terms of Lagrange multipliers. To do so, we consider the function $\mathcal{L} = f - \lambda g, $ where $f ( \mathbf{a} ) = \sum_{i=1}^{n} a_i b_i,$ and $g( \mathbf{a} ) = -1 + \sum_{i=1}^{n} a_i ^2.$

Appealing to the result on Lagrange multipliers, we know that if $f$ attains a max/min at a point $\mathbf{x}$ subject to the condition that $g( \mathbf{x} ) = 0,$ then $ \frac{ \partial \mathcal{L} }{ \partial x_i } = 0$ for every $i \in \{1,....., n \}$ at this point of extremum.

From these equations, we see that we must have $x_i = \frac{b_i}{ 2 \lambda }$ for each $i \in \{ 1,....,n \}.$

Now, using the constraint $|| \mathbf{x} || = 1,$ one can see that $\lambda \in \{ 1, -1 \}.$

It is easy to see that the case $\lambda = -1$ gives the point of constrained local min of $f$ and $\lambda = 1$ yields the point of constrained local max of $f.$

You can now put $\lambda = 1$ to obtain the version of Cauchy Schwarz inequality mentioned above. Of course this is equivalent to the original Cauchy Schwarz inequality (as justified above by homogenization).