# Demonstration of the inequality of Cauchy-Schwarz

After the demonstration of the inequality of Cauchy-Schwarz make by my professor, I still don't understand some steps of the demonstration.

To prove this inequality, my professor use the induction princile.

First, verify $P(1)$. \begin{align} \big(\sum^1{a_kb_k}\big)^2 &\le \big(\sum^1{a_k^2}\big)\big(\sum^1{b_k^2}\big) \\ \big(a_1b_1\big)^2 &= a_1^2b_1^2 \end{align}

We can also verify $P(2)$. \begin{align} \big(\sum^2{a_kb_k}\big)^2 &\le \big(\sum^2{a_k^2}\big)\big(\sum^2{b_k^2}\big) \\ \big(a_1b_1+a_2b_2\big)^2 &\le \big(a_1^2+a_2^2\big)\big(b_1^2+b_2^2\big) \\ a_1^2b_1^2+2a_1b_1a_2b_2+a_2^2b_2^2 &\le a_1^2b_1^2+a_1^2b_2^2+a_2^2b_1^2+a_2^2b_2^2 \\ 2a_1b_1a_2b_2 &\le a_1^2b_2^2+a_2^2b_1^2 \\ 0 &\le a_1^2b_2^2+a_2^2b_1^2 -2a_1b_1a_2b_2 \\ 0 &\le \big(a_1b_1-a_2b_2\big)^2 \end{align}

If we suppose $P(n)$ to be true, we can verify $P(n+1)$. \begin{align} \big(\sum^{n+1}{a_kb_k}\big)^2 &\le \big(\sum^{n+1}{a_k^2}\big)\big(\sum^{n+1}{b_k^2}\big) \\ \end{align}

By the inequality of the triangle, we can write the follow statement. $$\left|\sum^{n+1}{a_kb_k}\right| \le \left|\sum^{n}{a_kb_k}\right|+\left|a_{n+1}\right|\left|b_{n+1}\right|$$ So, we can assume, $$\left|\sum^{n+1}{a_kb_k}\right| \le \sqrt{\sum{a_k^2}}\sqrt{\sum{b_k^2}}+\left|a_{n+1}\right|\left|b_{n+1}\right|$$ because we know, $$\left|\sum{a_kb_k}\right| \le \sqrt{\sum{a_k^2}}\sqrt{\sum{b_k^2}}$$ If we define $A_1=\sqrt{\sum{a_k^2}}$, $B_2=\sqrt{\sum{b_k^2}}$, $A_2=\left|a_{n+1}\right|$ and $B_2=\left|b_{n+1}\right|$ we can write this inequation like this : $$\left|\sum^{n+1}{a_kb_k}\right| \le A_1B_1+A_2B_2$$ And than, there, it's where I don't understand... $$\left|\sum^{n+1}{a_kb_k}\right| \le \sqrt{A_1^2+A_2^2}\sqrt{B_1^2+B_2^2}$$

And why we have to verify $P(2)$.

• It's not true, as plugging almost any values into it will show. With $=$ replaced for $\leq$, this is the Cauchy-Schwarz inequality, essentially. – Thomas Andrews Jan 21 '15 at 22:13
• Yes. My mistake. But I still don't understand why. – hlapointe Jan 21 '15 at 22:14
• I think we need more context. – mick Jan 21 '15 at 22:14
• what are these steps that you don't understand?? – Surb Jan 21 '15 at 22:48
• Many nice proofs of Cauchy-Schwarz can be found on this thread: math.stackexchange.com/questions/436559/… – littleO Jan 21 '15 at 23:29

Actually, this expression is Cauchy-Schwarz's inequality itself. I'll give a proof and I hope you can follow. We have that $$(A_1+tB_1)^2 + (A_2+tB_2)^2 \geq 0$$ for all $t \in \Bbb R$, for being a sum of squares. This is equivalent to: $$(A_1^2+A_2^2)+2(A_1B_1+A_2B_2)t + (B_1^2+B_2^2)t^2 \geq 0.$$ This a polynomial in $t$ which is always positive, so $\Delta \leq 0$. Seeing this as $at^2+bt + c$, we have that $a = B_1^2+B_2^2, b = 2(A_1B_1+A_2B_2)$ and $c = A_1^2+A_2^2$, if this helps you seeing it. Then: $$\Delta = b^2-4ac = 4(A_1B_1+A_2B_2)^2-4(A_1^2+A_2^2)(B_1^2+B_2^2) \leq 0,$$ so that dividing by $4$ and taking roots, results $$|A_1B_1+A_2B_2| \leq \sqrt{A_1^2+A_2^2}\sqrt{B_1^2+B_2^2}.$$
Just notice that: $$(A_1^2+A_2^2)(B_1^2+B_2^2)-(A_1 B_1+A_2 B_2)^2 = (A_1 B_2 - A_2 B_1)^2 \geq 0.\tag{1}$$ $(1)$ is also known as Lagrange's identity.
• How can I use this identity to prove $A_1B_1+A_2B_2\le\sqrt{A_1^2+A_2^2}\sqrt{B_1^2+B_2^2}$? – hlapointe Jan 21 '15 at 22:18
• @hlapointe: $(1)$ proves $(A_1 B_1+A_2 B_2)^2 \leq (A_1^2+A_2^2)(B_1^2+B_2^2)$. – Jack D'Aurizio Jan 21 '15 at 22:19
My favorite proof: WLOG $\sum{a_k^2}=1$ and $\sum{b_k^2}=1$, then it's equivalent to showing $$\pm 2\sum{a_kb_k} \le 2\sqrt{\big(\sum{a_k^2}\big)\big(\sum{b_k^2}\big)}=\big(\sum{a_k^2}\big)+\big(\sum{b_k^2}\big),$$ which follows from $\pm 2ab\le a^2+b^2$.
One proof of (the real case) of the Cauchy-Schwarz inequality is via Lagrange's identity: $$\left( \sum_{i=1}^{n} a_{i}^{2} \right)\left( \sum_{i=1}^{n} b_{i}^{2} \right) = \left( \sum_{i=1}^{n} a_{i}b_{i} \right)^{2} + \sum_{i <j} (a_{i}b_{j} - a_{j}b_{i}) ^{2} .$$ There is a similar complex version too.