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

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  • $\begingroup$ 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 $\endgroup$ Commented Nov 12, 2022 at 12:57
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    $\begingroup$ math.stackexchange.com/questions/2319560/… $\endgroup$ Commented Nov 12, 2022 at 13:00
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    $\begingroup$ Do not remove your question after receiving answers to it, please $\endgroup$
    – postmortes
    Commented Nov 13, 2022 at 7:26

2 Answers 2

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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 $

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

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

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

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