Trích:
Nguyên văn bởi hong.qn  1. Cmr trong mọi tam giác $ABC $ ta có: $\boxed{\frac{1}{2Rr}\leqslant \frac{1}{a^2} +\frac{1}{b^2}+\frac{1}{c^2}} $ |
Ta có $S=\frac{abc}{4R}=pr \Rightarrow \frac{1}{2Rr}=\frac{a+b+c}{abc}=\frac{1}{ab}+\frac {1}{bc}+\frac{1}{ca} $
$\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}} \ge \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{1}{2R r} $
[RIGHT][I][B]Nguồn: MathScope.ORG[/B][/I][/RIGHT]