17. O inegalitate "tare" intr-un triunghi.

by Virgil Nicula, Apr 20, 2010, 3:09 AM

P1. Sa se arate ca intr-un triunghi $ABC$ exista inegalitatea "strong" $\boxed{\ \sum\ \frac{bc}{b+c-a}\ \le\ \frac{(5R-2r)(4R+r)^2}{4p(2R-r)}\ }$ .

Dem (Mateescu Constantin). Vom folosi relatia cunoscuta $\boxed {\ H\Gamma^2=4R^2\left[1-\frac{2p^2(2R-r)}{R(4R+r)^2}\right]\ }$ , unde $H$ este ortocentrul si $\Gamma$ este punctul lui Gergonne . Intrucat $H\Gamma^2\ \ge\ 0$ , din relatia precedenta deducem inegalitatea $\boxed {\ p^2\ \le\ \frac{R(4R+r)^2}{2(2R-r)}\ }\ \le\ 4R^2+4Rr+3r^2$ care este o intarire a inegalitatii Gerretsen. Deci $\sum\ \frac{bc}{b+c-a}=\sum\ \frac{4bc(p-b)(p-c)}{\prod (p-a)}=$ $\frac{4}{pr^2}\sum\ bc[p^2-p(b+c)+bc]$ .

Folosind relatiile $\sum b^2c^2=(p^2+r^2+4Rr)^2-16Rrp^2$ si $\sum bc(b+c)=\sum bc[(b+c+a)-a]=2p(p^2+r^2-2Rr)$ , ultima egalitate devine : $\boxed {\ \sum\ \frac{bc}{b+c-a}=\frac{p^2+(4R+r)^2}{2p}\ }$ .

Aplicand inegalitatea demonstrata anterior obtinem $:\ \boxed{\ \sum\ \frac{bc}{b+c-a}\ \le\ \frac{(5R-2r)(4R+r)^2}{4p(2R-r)}\ }\ \ \ \ \mbox{O.K.}$ .



P2.

Proof.
This post has been edited 18 times. Last edited by Virgil Nicula, Aug 21, 2017, 8:43 AM

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