1. ## Inequality help

I need help with the following inequality:

Show that $\displaystyle (a^2 - b^2)(c^2- d^2) \leq (ac-bd)^2$ for all $\displaystyle a,b,c,d$

Any help will be appreciated. Thank you.

2. Hi dude!

Originally Posted by MagicS06
I need help with the following inequality:

Show that $\displaystyle (a^2 - b^2)(c^2- d^2) \leq (ac-bd)^2$ for all $\displaystyle a,b,c,d$

Any help will be appreciated. Thank you.
$\displaystyle (a^2 - b^2)(c^2- d^2) \leq (ac-bd)^2$

<=>
[
you know what I did? First I solved
$\displaystyle (a^2 - b^2)(c^2- d^2) = - a^2d^2 + a^2c^2 + b^2d^2 - b^2c^2$
and then I solved
$\displaystyle (ac-bd)^2 = a^2c^2 - 2abcd + b^2d^2$
]

$\displaystyle - a^2d^2 + a^2c^2 + b^2d^2 - b^2c^2 \le a^2c^2 - 2abcd + b^2d^2$

step: -a^2c^2

$\displaystyle - a^2d^2 + b^2d^2 - b^2c^2 \le - 2abcd + b^2d^2$

step: -b^2d^2

$\displaystyle - a^2d^2 - b^2c^2 \le - 2abcd$

step: multiplied by (-1)

$\displaystyle a^2d^2 + b^2c^2 \ge 2abcd$

We know

$\displaystyle (ad-bc)^2 \ge 0$ (It is obvious, do you agree?)

and therefor it is

$\displaystyle (ad-bc)^2 = a^2d^2 -2adbc + b^2c^2 \ge 0$

step: +2adbc (this is equal to 2abcd)

=> $\displaystyle a^2d^2 + b^2c^2 \ge 2adbc$

q.e.d.

Any questions?

Best regards, Rapha