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.
Hi dude!
$\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