A acevipa Member Joined Sep 6, 2007 Messages 238 Gender Male HSC 2009 Sep 21, 2009 #1 Show that (a² - b²)(c²-d²) <= (ac-bd)^2
lychnobity Active Member Joined Mar 9, 2008 Messages 1,292 Gender Undisclosed HSC 2009 Sep 21, 2009 #2 acevipa said: Show that (a² - b²)(c²-d²) <= (ac-bd)^2 Click to expand... 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
acevipa said: Show that (a² - b²)(c²-d²) <= (ac-bd)^2 Click to expand... 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
O -Onlooker- Member Joined Mar 15, 2009 Messages 150 Gender Male HSC 2009 Sep 22, 2009 #3 lychnobity said: 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 Click to expand... Nice approach. + Rep.
lychnobity said: 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 Click to expand... Nice approach. + Rep.