13. Inegalitatea de la Sorin Borodi.

by Virgil Nicula, Apr 20, 2010, 1:38 AM

Lema. Intr-un triunghi $ ABC$ exista inegalitatea $ \boxed {\ 14R-r\ \ge \ 3s\sqrt 3\ }\ (1)$ (notatii standard).

Demonstratie. Inegalitatea $ s\sqrt 3\le 4R+r$ este cunoscuta (se arata usor). Asadar $ s\sqrt 3\le 4R+r\ \Longrightarrow$

$ 3s\sqrt 3\le 12R+3r=(14R-r)-2(R-2r)\ \le\ 14R-r\ \Longrightarrow\ 3s\sqrt 3\ \le 14R-r$ . Am folosit $ R\ \ge\ 2r$ .

PP1. Let $ ABC$ be a triangle in which denote the length $ l_a$ of the $ A$-bisector a.s.o. and $ a + b + c = 2s$ .

Prove that the following inequalities : $ \boxed {\ \begin{array}{cc} 1. & 3S\sqrt 3\\\\ 2. & s^2 - 8r(R - 2r)\end{array}\ \le\ l_a^2\ + \ l_b^2\ + \ l_c^2\ \le\ s^2\ }$ . (notatii standard).

Demonstratie. Vom folosi relatiile $ \left\|\begin{array}{c} ab+bc+ca=s^2+r(4R+r)\\\\ l^2_a=bc-\frac {a^2bc}{(b+c)^2}\end{array}\right\|$ si inegalitatea Gerretsen $ \boxed {\ s^2\ge 16Rr-5r^2\ }\ (2)$ .

Asadar $ (b+c)^2\ge 4bc$ etc $ \Longrightarrow\ \sum l_a^2=\sum bc-\sum\frac {a^2bc}{(b+c)^2}$ $ \ge \sum bc-\sum\frac {a^2}{4}=\sum bc-\frac 14\cdot\left[\left(\sum a\right)^2-2\sum bc\right]=$

$ s^2+r(4R+r)-\frac 14\cdot\left\{4s^2-2\left[s^2+r(4R+r)\right]\right\}=$ $ s^2+r(4R+r)-\frac 12\cdot \left[s^2-r(4R+r)\right]=\frac 12\cdot\left[s^2+3r(4R+r)\right]\ \stackrel{(2)}{\ge}$

$ \frac 12\cdot\left[16Rr-5r^2+3r(4R+r)\right]=r(14R-r)\ \stackrel{(1)}{\ge}\ r\cdot 3s\sqrt 3=3S\sqrt 3$ . In concluzie $ l_a^2+l_b^2+l_c^2\ \ge\ 3S\sqrt 3\ >\ 5S$ .
This post has been edited 3 times. Last edited by Virgil Nicula, Nov 23, 2015, 7:01 AM

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