proof of Cauchy-Schwarz inequality for real numbers

The version of the Cauchy-Schwartz inequality we want to prove is

$${\left(\sum _{k=1}^n{a}_k{b}_k\right)}^2\le \sum _{k=1}^n{a}_k^2\cdot \sum _{k=1}^n{b}_k^2,$$

where the $a_k$ and $b_k$ are real numbers, with equality holding only in thecase of proportionality, $a_k=\lambda b_k$ for some real $\lambda$ for all $k$.

The proof is by direct calculation:

${\displaystyle \sum _{k=1}^n}{a}_k^2\cdot {\displaystyle \sum _{k=1}^n}{b}_k^2-{\left({\displaystyle \sum _{k=1}^n}{a}_k{b}_k\right)}^2$$={\displaystyle \sum _{k,l=1}^n}{a}_k^2{b}_l^2-a_k{b}_k{a}_l{b}_l$

$={\displaystyle \sum _{k,l=1}^n}{\displaystyle \frac{1}{2}}(a_k^2{b}_l^2+a_l^2{b}_k^2)-(a_k{b}_l)(a_l{b}_k)$

$={\displaystyle \frac{1}{2}}{\displaystyle \sum _{k,l=1}^n}{(a_k{b}_l)}^2-2(a_k{b}_l)(a_l{b}_k)+{(a_l{b}_k)}^2$

$={\displaystyle \frac{1}{2}}{\displaystyle \sum _{k,l=1}^n}{(a_k{b}_l-a_l{b}_k)}^2$

$\ge 0.$

The above identity implies that the Cauchy-Schwarz inequality holds.Moreover, it is an equality only when

$$a_k{b}_l-a_l{b}_k=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}\frac{a_k}{b_k}=\frac{a_l}{b_l}\text{ or }\frac{b_k}{a_k}=\frac{b_l}{a_l}\text{ or }{a}_k=b_k=0,$$

for all $k$ and $l$. In other words, equality holds only when $a_k=\lambda b_k$ for all $k$ for some real number $\lambda$.