Inegalitate geometrica

[nica]

Sa se arate ca in orice triunghi are loc inegalitatea: \( R+r\geq\frac{\ r_ar_b+r_br_c+r_cr_a}{r_a+r_b+r_c} \)

[Mateescu Constantin]

Stim ca \( r_ar_b+r_br_c+r_cr_a=p^2 \) si \( r_a+r_b+r_c=4R+r \) .

Astfel inegalitatea devine: \( R+r\ge \frac{p^2}{4R+r}\ \Longleftrightarrow\ 4R^2+5Rr+r^2\ge p^2 \).

Dar din Euler si Gerretsen avem \( 4R^2+5Rr+r^2\ge 4R^2+4Rr+3r^2\ge p^2 \) .