Inegalitate geometrica simpla in triunghi

[mihai miculita]

Aratati ca intr-un triunghi oarecare $ABC,$ avem inegalitatea: $\dfrac{a^2}{r_br_c}+\dfrac{b^2}{r_ar_c}+\dfrac{c^2}{r_ar_b}\ge 4.$

[Andi Brojbeanu]

Inegalitatea se poate demonstra in mai multe moduri:
i) Cu inegalitatea C.B.S. rezulta $\displaystyle \sum{\frac{a^2}{r_br_c}}\ge \frac{(a+b+c)^2}{r_ar_b+r_br_c+r_cr_a}=\frac{4p^2}{\sum{\frac{S^2}{(p-b)(p-c)}}}=$$\displaystyle \frac{4p^2}{p^2r^2}\cdot \frac{(p-a)(p-b)(p-c)}{(p-a)+(p-b)+(p-c)}=\frac{4}{r^2}\cdot \frac{pr^2}{p}=4$.
ii) Cu inegalitatea mediilor rezulta $\displaystyle \sum{\frac{a^2}{r_br_c}}\ge 3\sqrt[3]{\frac{a^2b^2c^2}{r_a^2r_b^2r_c^2}}=3\left(\frac{4pRr}{p^2r}\right)^{\frac{2}{3}}=3\left(\frac{4R}{p}\right)^{\frac{2}{3}}\ge 3\left(\frac{8}{3\sqrt{3}}\right)^{\frac{2}{3}}$$\displaystyle =3\cdot \frac{4}{3}=4$. Am folosit inegalitatea lui Mitrinovic $\displaystyle \frac{3\sqrt{3}}{2}R\ge p$ si faptul ca $\displaystyle r_ar_br_c=\frac{S^3}{(p-a)(p-b)(p-c)}=\frac{p^3r^3}{pr^2}=p^2r$.
iii) Se poate demonstra usor ca $\displaystyle \sum{\frac{a^2}{r_br_c}}=4\left(\frac{R}{r}-1\right)$, de unde rezulta concluzia folosind inegalitatea lui Euler $\displaystyle R\ge 2r$.