mateforum.ro

Inegalitatea Weizenbock intr-un triunghi oarecare ABC:
\( a^2 + b^2 + c^2 \ge 4S\sqrt 3 \)

Demonstratia 1:
\( \begin{array}{l}
\left. \begin{array}{l}
{\rm ctgA = }\frac{{{\rm cosA}}}{{{\rm sinA}}} = \frac{{\frac{{b^2 + c^2 - a^2 }}{{2bc}}}}{{\frac{a}{{2R}}}} = \frac{{R(b^2 + c^2 - a^2 )}}{{abc}} \Rightarrow ctgA + ctgB + ctgC = \frac{{a^2 + b^2 + c^2 }}{{4S}} \\
ctgA + ctgB + ctgC \ge \sqrt 3 \\
\end{array} \right| \Rightarrow a^2 + b^2 + c^2 \ge 4S\sqrt 3 \\
\end{array}
\)
Demonstratia 2:
Fie T punctul in care m(ATB)=m(BTC)=m(ATC)=120 si AT=x, BT=y, CT=z. Atunci
\( \begin{array}{l}
a^2 = y^2 + z^2 - 2yz\cos 120 = y^2 + z^2 + yz; \\
b^2 = z^2 + x^2 + xz; \\
c^2 = x^2 + y^2 + xy; \\
S = (BTC) + (CTA) + (ATB) = \frac{1}{2}yz\sin 120 + \frac{1}{2}zx\sin 120 + \frac{1}{2}xy\sin 120 = \frac{{\sqrt 3 }}{4}(xy + yz + zx) \Rightarrow 4S\sqrt 3 = 3(xy + yz + zx) \\ \end{array} \) de unde rezulta imediat concluzia.
Pentru demonstratia (2) vezi si cartea Inegalitati matematice (modele inovatoare).

Se poate demonstra o inegalitate mai tare :

\( a^2+b^2+c^2\ge 4\sqrt{3}S+(a-b)^2+(b-c)^2+(c-a)^2 \) (Hadwiger-Finsler)

alex2008 wrote:Se poate demonstra o inegalitate mai tare :

\( a^2+b^2+c^2\ge 4\sqrt{3}S+(a-b)^2+(b-c)^2+(c-a)^2 \) (Hadwiger-Finsler)

Scriem inegalitatea sub forma
\( a^2-(b-c)^2+b^2-(c-a)^2+c^2-(a-b)^2\ge 4S\sqrt{3} \)
\( \Longleftrightarrow (a-b+c)(a+b-c)+(b-c+a)(b+c-a)+(c-a+b)(c+a-b)\ge 4S\sqrt{3} \)

Notam \( x=-a+b+c \) si analoagele \( (x,\ y,\ z>0) \)
\( \Longrightarrow 4S\sqrt{3}=\sqrt{3(x+y+z)xyz} \)

Astfel obtinem: \( xy+yz+zx\ge \sqrt{3(x+y+z)xyz \)

Impartind prin \( xyz \) si notand \( m=\frac{1}{x},\ n=\frac{1}{y},\ p=\frac{1}{z} \) obtinem

\( \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge \sqrt{3\left\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right\)} \ \Longleftrightarrow m+n+p\ge \sqrt{3(mn+np+pm)} \), inegalitate cunoscuta.

Ambele inegalitati, atat Weitzenböck cat si Hadwiger-Finsler se pot gasi aici alaturi de demonstratiile lor -> http://compactorange.googlepages.com/tin2006new.pdf

Daca tot s-a vorbit despre acest tip de inegalitati sa o menitonam
si pe cea mai tare dintre ele (cu toate ca tehnica de rezolvare nu este deloc grea)
si anume problema 18 de la Sharygin 2008 propusa de A. Abdullayev:
http://www.mathlinks.ro/viewtopic.php?p=1245889#1245889.