proof of Cauchy-Schwarz inequality

If $a$ and $b$ are linearly dependent, we write $𝒃=\lambda 𝒂$. So we get:

$${⟨𝒂,\lambda 𝒂⟩}^2={\lambda }^2{⟨𝒂,𝒂⟩}^2={\lambda }^2{||𝒂||}^4={||𝒂||}^2{||𝒃||}^2.$$

So we have equality if $𝒂$ and $𝒃$ are linearly dependent. In the other case we look at the quadratic function

$${||x\cdot 𝒂+𝒃||}^2=x^2{||𝒂||}^2+2x⟨𝒂,𝒃⟩+{||𝒃||}^2.$$

This function is positive for every real $x$, if $𝒂$ and $𝒃$ are linearly independent. Thus it has no real zeroes, which means that

$${⟨𝒂,𝒃⟩}^2-{||𝒂||}^2{||𝒃||}^2$$

is always negative. So we have:

which is the Cauchy-Schwarz inequality if $𝒂$ and $𝒃$ are linearly independent.