# A natural proof of the Cauchy-Schwarz inequality

Most of the proofs of the Cauchy-Schwarz inequality on a pre-Hilbert space use a fact that if a quadratic polynomial with real coefficients takes positive values everywhere on the real line, then its discriminant is negative(e.g. Conway: A course in functional analysis). I think this is somewhat tricky. Moreover I often forget its proof when the pre-Hilbert space is defined over the field of complex numbers. Is there a more natural proof (hence it's easy to remember) which is based on a completely different idea?

• To someone who voted to close, would you please explain the reason for the vote? – Makoto Kato Jul 5 '13 at 5:59
• You might want to browse the first chapter of The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities by J. Michael Steele. – Did Jul 5 '13 at 6:26
• I'm not sure about what could qualify as "natural" or "more natural" proof for you, but in this case I can't think, off the top of my head, of anything *simpler", basic and straightforward as working with a quadratic's discriminant: this is Junior High School stuff! – DonAntonio Jul 5 '13 at 8:45
• @DonAntonio As I wrote, I don't think the complex pre-Hilbert space case is not so straightforward. – Makoto Kato Jul 5 '13 at 9:01
• A question which has 3 upvotes, 2 favorites and a 6 upvoted answer should not be closed. Please reset the close votes. – Makoto Kato Jul 5 '13 at 19:24

Recall the Pythagorean theorem: If $u_1, \cdots u_n$ are pairwise orthogonal, then $$\| u_1 + \cdots u_n \|^2 = \|u_1\|^2 + \cdots + \| u_n \|^2.$$

I want to use this to tell us something about two non-zero vectors $u$ and $v,$ but they aren't necessarily orthogonal. So consider the projection of $u$ onto the plane of vectors orthogonal to $v:$ $$w = u - \frac{ \langle u,v \rangle}{\|v\|^2} v.$$ This is certainly orthogonal to $v,$ and the Pythagorean theorem applied to $w$ and $v$ gives the Cauchy-Schwarz inequality.

• This is excellent. Thanks. By the way, I think $<v, u>$ is a typo($<v, v>$). – Makoto Kato Jul 5 '13 at 9:19
• @MakotoKato Thank you for spotting that. Also, I find using \langle , \rangle for inner products is more pleasing to the eye than $<$ and $>.$ – Ragib Zaman Jul 5 '13 at 14:14
• @RagibZaman You're welcome. I just didn't know how to write $\langle, \rangle$ instead of < and >. By the way again, I think $\|v||$ should be squared or replaced by $\langle v, v \rangle$. Regards. – Makoto Kato Jul 5 '13 at 19:19
• I fail to see how you obtain the CS inequality in your last paragraph. Would you explain that? – Martin Argerami Jul 5 '13 at 20:42
• @MartinArgerami Let $u_1 = w$, $u_2 = \frac{ \langle u,v \rangle}{\langle v, v \rangle} v$. Then apply the Pythagorean formula $\| u_1 + u_2\|^2 = \|u_1\|^2 + \| u_2 \|^2 \ge \|u_2\|^2$. Regards. – Makoto Kato Jul 6 '13 at 0:41

There is also an approach by "amplification" which is really cool. Also the exact same trick works to prove Hölder's inequality and is generally a very important principle for improving inequalities.

It goes like this: We start out with $$\langle a-b,a-b\rangle\ge 0$$ for $a,b$ in your inner product space, and $a\not=0$, $b\not=0$. This implies $$2\langle a,b\rangle\le \langle a, a\rangle + \langle b, b\rangle$$ Now notice that the left hand side is invariant under the scaling $a\mapsto \lambda a$, $b\mapsto \lambda^{-1}b$ for $\lambda>0$. This gives $$2\langle a,b\rangle \le \lambda^2 \langle a,a\rangle + \lambda^{-2}\langle b, b\rangle$$ Now look at the right hand side as a function of the real variable $\lambda$ and find the optimal value for $\lambda$ using calculus (set the derivative to $0$):

$$\lambda^2=\sqrt{\frac{\langle b,b\rangle}{\langle a,a\rangle}}$$

Plugging this value in, we obtain

$$2\langle a,b\rangle\le \sqrt{\langle a,a\rangle}\sqrt{\langle b,b\rangle}+\sqrt{\langle a,a\rangle}\sqrt{\langle b,b\rangle}$$

i.e.

$$\langle a,b\rangle\le\sqrt{\langle a,a\rangle}\sqrt{\langle b,b\rangle}$$

Notice how we took a trivial observation and "optimized" the expression by exploiting scaling invariance.

• As explained (and expanded) by Terence Tao on his blog. – Did Jul 5 '13 at 13:08
• Could your proof be modified to show that the equality holds iff $a$ is proportional to $b$? I've been thinking a while on it, but I haven't figured anything out. – jinawee Oct 1 '14 at 19:54
• @jinawee: Yes, the only inequality in this proof is $\langle \mu a-b,\mu a-b\rangle\ge 0$ for an optimal constant $\mu$ depending on $a,b$. Equality implies $b=\mu a$. – J.R. Oct 2 '14 at 7:17

The inequality $| \langle a, b \rangle | \leq \| a \| \| b \|$ can be rewritten as $$| \langle a, \frac{b}{\|b\|} \rangle | \leq \| a \|$$ (assuming $b \neq 0$).

On the left we see the component of $a$ on the unit vector $u = \frac{b}{\| b \|}$. On the right we have the norm of $a$. Of course the norm of $a$ is larger than the component of $a$ on $u$, this is very intuitive. To make this into a rigorous proof, we merely have to write $a = \langle a, u \rangle u + v$, note that $v \perp u$, and use the pythagorean theorem.

Another approach is to start with the inequality $$0 \leq \| a - \text{proj}_b(a) \|^2.$$

Just expand the right hand side and Cauchy-Schwarz pops out.

This is not exactly an answer but an explanation for the idea behind the Ragib Zaman's answer.

Let $K$ be the field of real numbers or the field of complex numbers. Let $E$ be a pre-Hilbert space over $K$.

Let $x, y$ be elements of $E$ such that $\langle x, y \rangle = 0$. Then $\|x + y\|^2 = \langle x+y, x+ y\rangle = \langle x, x\rangle + \langle x, y \rangle + \langle y, x\rangle + \langle y,y\rangle = \|x\|^2 + \|y\|^2$.

Let $x, y$ be elements of $E$ such that $\langle x - y, y \rangle = 0$. Then, by the above formula, $\|x\|^2 = \|x - y\|^2 + \|y\|^2$. Hence $\|x\| \ge \|y\|$.

Finally let $u, v$ be non-zero elements of $E$. Let $t$ be an element of $K$ such that $\langle u - tv, v \rangle = 0$. $t$ must be $\frac{ \langle u,v \rangle}{\langle v, v\rangle}$ since $v \neq 0$. Then $\|u\| \ge \|tv\|$ by the previous inequality. This leads to the Cauchy-Schwarz inequality immediately.

Here's my favorite proof, mainly because it's nicely symmetric, easy to remember and not impossible to come up with (the main trick is that $2=1+1$):

We want to prove $$|\langle x,y\rangle|\le\|x\|\|y\|$$ This is linear in $x$ or $y$ and obviously holds for $x=0$ or $y=0$. Therefore without loss of generality we can suppose that $\|x\|=\|y\|=1$: $$|\langle x,y\rangle|\le\|x\|\|y\|\Longleftrightarrow|\langle x,y\rangle|\le1\Longleftrightarrow2|\langle x,y\rangle|\le2\Longleftrightarrow2|\langle x,y\rangle|=\|x\|^2+\|y\|^2$$

Let $u\in\mathbb C$ be a complex unit (i.e. $|u|=1$), then \begin{align}0\le\|x-uy\|^2&=\langle x-uy,x-uy\rangle=\langle x,x\rangle-u\langle x,y\rangle-\overline{u\langle x,y\rangle}+\langle y,y\rangle\\&=\|x\|^2+\|y\|^2-2\,\text{Re}(u\langle x,y\rangle)\end{align} Now just set $u$ so that $\text{Re}(u\langle x,y\rangle)=|\langle x,y\rangle|$ and you are done.

Of course in the real case you can just expand $0\le\|x\pm y\|^2$.