Inegalitate intr-un triunghi
Posted: Sun Jun 08, 2008 10:20 am
Daca \( a,b,c \) sunt lungimile laturilor unui triunghi iar \( p \) semiperimetrul sau, sa se demonstreze ca \( \frac{a^2}{p-a}+\frac{b^2}{p-b}+\frac{c^2}{p-c} \geq 4p \)
Page 1 of 1
Posted: Sun Jun 08, 2008 11:00 am
Se aplica Cauchy o singura data...si gata.
Posted: Sat Jun 21, 2008 7:39 pm
\( \sum\frac{a^2}{p-a}\cdot\sum(p-a)\ge(\sum a)^2\Rightarrow\sum\frac{a^2}{p-a}\ge\frac{(\sum a)^2}{\sum(p-a)}=\frac{(2p)^2}{3p-a-b-c}=\frac{4p^2}{p}=4p \)