Egalitate
Moderators: Bogdan Posa, Laurian Filip
Egalitate
Se stie ca \( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c} \) . Sa se arate ca \( \frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{a^3+b^3+c^3} \) .
. A snake that slithers on the ground can only dream of flying through the air.
\( \displaystyle\frac{1}{a}+\frac{1}{c}=\frac{1}{a+b+c}-\frac{1}{c}\Leftrightarrow \frac{a+b}{ab}=-\frac{a+b}{c(a+b+c)}\Leftrightarrow (a+b)(a+c)(b+c)=0 \)
Rezulta \( a=-b \) sau \( a=-c \) sau \( b=-c \)
Presupunem ca \( a=-b \)
Atunci \( \frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{a^3}-\frac{1}{a^3}+\frac{1}{c^3}=\frac{1}{c^3} \)
iar \( \frac{1}{a^3+b^3+c^3}=\frac{1}{a^3-a^3+c^3}=\frac{1}{c^3} \)
Obs. Concluzia poate fi generalizata la \( \frac{1}{a^{2n+1}}+\frac{1}{b^{2n+1}}+\frac{1}{c^{2n+1}}=\frac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}, \ \forall n\in\mathbf{N} \)
Rezulta \( a=-b \) sau \( a=-c \) sau \( b=-c \)
Presupunem ca \( a=-b \)
Atunci \( \frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{a^3}-\frac{1}{a^3}+\frac{1}{c^3}=\frac{1}{c^3} \)
iar \( \frac{1}{a^3+b^3+c^3}=\frac{1}{a^3-a^3+c^3}=\frac{1}{c^3} \)
Obs. Concluzia poate fi generalizata la \( \frac{1}{a^{2n+1}}+\frac{1}{b^{2n+1}}+\frac{1}{c^{2n+1}}=\frac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}, \ \forall n\in\mathbf{N} \)