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 $P(2)$.
 A: 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.
A: 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 $$|A_1B_1+A_2B_2| \leq \sqrt{A_1^2+A_2^2}\sqrt{B_1^2+B_2^2}.$$
A: 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 $\pm 2ab\le a^2+b^2$.
A: 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.
