Let $b$ and $d$ be any two $p\times 1$ vectors. Show that $$(b'd)^2\leq (b'b)(d'd)$$ with equality if and only if $b=cd$.
I get this proof from the book.
Let $b-xd$ where $x$ is an scalar. Since the length of $b-xd>0$ for $b-xd\neq 0$
$$0<(b-xd)'(b-xd)=b'b-xd'b-b'xd+x^2d'd$$ $$=b'b-2x(b'd)+x^2(d'd)$$
since the expression is a quadratic form in $x$ $$0<b'b-\frac{(b'd)^2}{d'd}+\frac{(b'd)^2}{d'd}-2x(b'd)+x^2(d'd)$$ $$=b'b-\frac{(b'd)^2}{d'd}+(d'd)\Big(x-\frac{b'd}{d'd}\Big)^2$$
choosing $x=\frac{b'd}{d'd}$
$$0<b'b-\frac{(b'd)^2}{d'd}$$
Then $(b'd)^2<(b'b)(d'd)$ if $b\neq xd$ for some $x$. Note that if $b=cd$ then $0=(b-cd)'(b-cd)$ and the same argument produces $(b'd)^2=(b'b)(d'd)$
Is there any proof more intuitive? I didn't get the idea of using $b-xd$ to proof. In the last steps I see that work, but I don't understood what is the idea behind.