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Posted: **Tue Jun 15, 2010 2:32 am**

Fie triunghiul \( ABC \) cu incercul \( C(I) \) si circumcercul \( w \) . Consideram trei puncte \( D\in (BC) \) , \( E\in (CA) \) ,

\( F\in (AB) \) pentru care notam \( \{A,X\}=AD\cap w \) , \( \{B,Y\}=BE\cap w \) , \( \{C,Z\}=CF\cap w \) . Sa se arate ca

\( \frac {XA}{XD}+\frac {YB}{YE}+\frac {ZC}{ZF}\ \ge\ \left(\frac {b+c}{a}\right)^2+\left(\frac {c+a}{b}\right)^2+\left(\frac {a+b}{c}\right)^2 \) cu egalitate daca si numai daca \( I\ \in\ AD\cap BE\cap CF \) .

Posted: **Tue Jun 15, 2010 3:22 am**

Vom arata ca are loc inegalitatea : \( \frac{XA}{XD}\ \ge\ \left(\frac {b+c}a\right)^2 \) , cu egalitate cand \( AD \) este bisectoarea \( \widehat {BAC} \) .

Scriem relatia Stewart in \( \triangle\ ABC \) pentru ceviana \( \begin{array}{ccc}AD & : & a\cdot AD^2 & + & a\cdot BD\cdot CD & = & c^2\cdot CD & + & b^2\cdot BD\end{array} \)

\( \begin{array}{cccc}\Longrightarrow & \frac{AD^2}{BD\cdot CD} & + & 1 & = & \frac{c^2}{a\cdot BD} & + & \frac{b^2}{a\cdot CD} & \stackrel{\small C.B.S.}{\ge} & \left(\frac{b+c}a\right)^2\end{array} \) , cu egalitate iff \( \frac{c}{a\cdot BD}=\frac{b}{a\cdot CD}\ \Longleftrightarrow\ AD \) bisectoare .

Insa cum \( X\in AD\cap w\ ,\ A\ne X\ \Longrightarrow\ BD\cdot CD=AD\cdot DX \) , si astfel ultima inegalitate va fi echivalenta cu :

\( \frac{AD^2+AD\cdot DX}{AD\cdot DX}\ \ge\ \left(\frac {b+c}a\right)^2\ \Longleftrightarrow\ \fbox{\ \frac{XA}{XD}\ \ge\ \left(\frac {b+c}a\right)^2\ } \) . Evident, insumarea relatiilor analoage ofera concluzia .

mateforum.ro

O inegalitate intr-un triunghi si cercul sau circumscris.

O inegalitate intr-un triunghi si cercul sau circumscris.