Inegalitatea 5, trigonometrica, cu (1-cos A) si cosA

Moderators: Laurian Filip, Beniamin Bogosel, Filip Chindea

Post Reply
User avatar
Cezar Lupu
Site Admin
Posts: 612
Joined: Wed Sep 26, 2007 2:04 pm
Location: Bucuresti sau Constanta
Contact:

Inegalitatea 5, trigonometrica, cu (1-cos A) si cosA

Post by Cezar Lupu »

Sa se arate ca in orice triunghi \( ABC \) este adevarata urmatoarea inegalitate

\( (1-cos A)(1-\cos B)(1-\cos C)\geq\ cos A\cos B\cos C \).
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
User avatar
Filip Chindea
Newton
Posts: 324
Joined: Thu Sep 27, 2007 9:01 pm
Location: Bucharest

Post by Filip Chindea »

Pur si simplu facem evaluarea din T. Cosinusului:
\( \Leftrightarrow \prod \left( 1 - \frac{-a^2 + b^2 + c^2}{2bc} \right) \ge \prod \frac{-a^2 + b^2 + c^2}{2bc} \) \( \Leftrightarrow \prod \frac{(a - b + c)(a + b - c)}{2bc} \ge \frac{\prod (-a^2 + b^2 + c^2)}{8a^2b^2c^2} \) \( \Leftrightarrow \prod (-a + b + c)^2 \ge \prod (-a^2 + b^2 + c^2) \). Acum simplificam problema prin substitutia \( a = y + z \), etc., \( x, y, z > 0 \), si ramâne \( 8x^2y^2z^2 \ge \prod (x^2 + xy + xz - yz) \) \( \Leftrightarrow \sum_{sym} x^4y^2 + 2 \sum x^3y^3 + 6x^2y^2z^2 \ge 2 \sum x^4yz + 2 \sum_{sym} x^3y^2z \), ceea ce rezulta din \( \sum_{sym}x^4y^2 = \sum x^4(y^2 + z^2) \ge 2 \sum x^4yz \) (AM-GM) si din \( \sum x^3y^3 + 3x^2y^2z^2 \ge \sum_{sym} x^3y^2z \), care este chiar Schur de gradul 3 in care am substituit variabilele cu \( yz, zx \) si \( xy \).
Life is complex: it has real and imaginary components.
Post Reply

Return to “Clasa a IX-a”