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

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

• Welcome. Geometrically, your inequality is interpreted as $\forall \theta\in \Bbb R, |\sin\theta|\leq 1$. I believe you should do geometry rather than discrete math here Nov 12, 2022 at 12:57
• math.stackexchange.com/questions/2319560/… Nov 12, 2022 at 13:00
• Do not remove your question after receiving answers to it, please Nov 13, 2022 at 7:26

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

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

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

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

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

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

• This solution is eloquent(+1) but over-kill at the same time. See the answer of @scarface Nov 12, 2022 at 13:06