# A Nifty Proof of the Cauchy-Schwarz Inequality

This blog post is entirely based on the start of this blog post by Terry Tao. I highly recommend reading the post. It gives an interesting insight into how Terry sometimes thinks about proving inequalities. He also gives a number of cool and more substantial examples.

The main idea in the blog post is that Terry likes to do “arbitrage” on an inequality to improve it. By starting with a weak inequality he exploits the symmetry of the environment he is working in to get better and better inequalities. He first illustrates this with a proof of the Cauchy-Schwarz inequality. The proof given is really nifty and much more memorable than previous proofs I’ve seen. I felt that just had to write it up and share it.

Let $(V,\langle, \rangle)$ be an inner product space. The Cauchy-Schwarz inequality states that for all $v,w \in V$, $|\langle v, w \rangle | \le \|v\| \|w\|$. It’s an important result that leads, among other things, to a proof that $\| \cdot \|$ satisfies the triangle inequality. There are many proofs of the Cauchy-Schwarz inequality but here is the one Terry presents.

Since $\langle \cdot, \cdot \rangle$ is positive definite we have $0 \le \langle v-w,v-w\rangle$. Now using the fact that $\langle \cdot, \cdot \rangle$ is additive in each coordinate we have

$0 \le \langle v,v \rangle -\langle v, w \rangle -\langle w,v\rangle+ \langle w,w \rangle$.

Since $\langle w,v\rangle = \overline{\langle v,w\rangle}$, we can rearrange the above expression to get the inequality

$\text{Re}(\langle v,w \rangle) \le \frac{1}{2}\left(\| v \|^2 + \|w\|^2\right)$.

And now it is time to exploit the symmetry of the above expression and turn this inequality into the Cauchy-Schwarz inequality. The above inequality is worse than the Cauchy Schwarz inequality for two reasons. Firstly, unless $\langle v, w \rangle$ is a positive real number, the left hand side is smaller than $|\langle v,w \rangle|$. Secondly, unless $\|v\|=\|w\|$, the right hand side is larger than the quantity $\|v\|\|w\|$ that we want. Indeed we want the geometric mean of $\|v\|^2$ and $\|w\|^2$ whereas we currently have the arithmetic mean on the right.

Note that the right hand side is invariant under the symmetry $v \mapsto e^{i \theta} v$ for any real number $\theta$. Thus choose $\theta$ to be the negative of the argument of $\langle v,w \rangle$. This turns the left hand side into $|\langle v,w \rangle |$ while the right hand side remains invariant. Thus we have done our first bit of arbitrage and now have the improved inequality

$| \langle v,w \rangle | \le \frac{1}{2}\left(\|v\|^2 + \|w\|^2\right)$

We now turn our attention to the right hand side and observe that the left hand side is invariant under the map $(v,w) \mapsto \left(c\cdot v, \frac{1}{c} \cdot w\right)$ for any $c > 0$. Thus by choosing $c$ we can minimize the right hand side. A little bit of calculus shows that the best choice is $c = \sqrt{\frac{\|w\|}{\|v\|}}$ (this is valid provided that $v,w \neq 0$, the case when $v=0$ or $w=0$ is easy since we would then have $\langle v,w \rangle=0$). If we substitute in this optimal value of $c$, the right hand side of the above inequality becomes

$\frac{1}{2}\left( \left\| \sqrt{\frac{\|w\|}{\|v\|}}\cdot v \right \|^2 +\left \| \sqrt{\frac{\|v\|}{\|w\|}} \cdot w\right \|^2 \right)=\frac{1}{2}\left(\frac{\|w\|}{\|v\|}\|v\|^2+\frac{\|v\|}{\|w\|}\|w\|^2 \right) = \|v\|\|w\|.$

Thus we have turned our weak starting inequality into the Cauchy-Schwarz inequality! Again I recommend reading Terry’s original post to see many more examples of this sort of arbitrage and symmetry exploitation.