Inequalities Question (1 Viewer)

[acevipa]

Show that (a² - b²)(c²-d²) <= (ac-bd)^2

 

[lychnobity]

Show that (a² - b²)(c²-d²) <= (ac-bd)^2

Expand both sides

a²c² - a²d² - b²c² + b²d² ≤ a²c² - 2abcd + b²d²

- a²d² - b²c² ≤ -2abcd

a²d² + b²c² ≥ 2abcd

.'. to prove (a² - b²)(c² - d²) ≤ (ac-bd)², is to prove that:

a²d² + b²c² ≥ 2abcd

(ad - bc)² ≥ 0

a²d² + b²c² - 2abcd ≥ 0

.'. a²d² + b²c² ≥ 2abcd

.'. (a² - b²)(c² - d²) ≤ (ac-bd)²

 

[-Onlooker-]

Expand both sides

a²c² - a²d² - b²c² + b²d² ≤ a²c² - 2abcd + b²d²

- a²d² - b²c² ≤ -2abcd

a²d² + b²c² ≥ 2abcd

.'. to prove (a² - b²)(c² - d²) ≤ (ac-bd)², is to prove that:

a²d² + b²c² ≥ 2abcd

(ad - bc)² ≥ 0

a²d² + b²c² - 2abcd ≥ 0

.'. a²d² + b²c² ≥ 2abcd

.'. (a² - b²)(c² - d²) ≤ (ac-bd)²

Nice approach.

+ Rep.