mateforum.ro

Sa se arate ca in orice triunghi avem :

\( (p-a)^3(p-c)+(p-b)^3(p-a)+(p-c)^3(p-b)\ge S^2 \)

Stefan Smarandache G.M.

Impartind intai prin \( (p-a)(p-b)(p-c) \) si apoi prin \( p \), inegalitatea data devine:
\( \frac{\left(p-a\right)^{2}}{p-b}+\frac{\left(p-b\right)^{2}}{p-c}+\frac{\left(p-c\right)^{2}}{p-a}\ge\frac{S^{2}}{\left(p-a\right)\left(p-b\right)\left(p-c\right)}\Leftrightarrow\frac{1}{p}\left(\frac{\left(p-a\right)^{2}}{p-b}+\frac{\left(p-b\right)^{2}}{p-c}+\frac{\left(p-c\right)^{2}}{p-a}\right)\ge\frac{S^{2}}{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}=1 \).

Aceasta ultima inegalitate este adevarata, deoarece conform inegalitatii Cauchy-Buniakowski-Schwarz avem:

\( \frac{1}{p}\left(\frac{\left(p-a\right)^{2}}{p-b}+\frac{\left(p-b\right)^{2}}{p-c}+\frac{\left(p-c\right)^{2}}{p-a}\right)\ge\frac{1}{p}\cdot\frac{\left(3p-\left(a+b+c\right)\right)^{2}}{3p-\left(a+b+c\right)}=\frac{1}{p}\left(3p-\left(a+b+c\right)\right)=\frac{1}{p}\cdot p=1 \).