Discrete Mathematics: If $a,b,c,d \in \mathbb{R}$, then $(ab-cd)^2 \leq (a^2+c^2)(b^2+d^2)$.

Question

How would I prove this? I started by expanding the terms, but afterwards I am not sure what more to do to proceed.

Answer 1

Let's use your method. Let's expand the terms and we get

$a^2b^2 + c^2d^2 -2abcd \leq a^2b^2 + a^2d^2 + b^2c^2 + c^2d^2$

$\iff -2abcd \leq a^2d^2 + b^2c^2 $

$\iff a^2d^2 + b^2c^2 +2abcd \geq 0 $

$\iff (ad + bc)^2 \geq 0$.

Answer 2

Apply the Cauchy-Schwarz inequality to the pair of vectors $(a,c)$ and $(b,-d)$ (assuming the standard inner product on $\mathbb{R}^2$).