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O dificila inegalitate geometrica (Own, CRUX).
Posted: Sun Sep 30, 2007 2:27 pm
by Virgil Nicula
Sa se arate ca intr-un triunghi \( ABC \) exista inegalitatea \( \frac ba+\frac {a}{b+c}\ge\sqrt {2+\frac {\left|b^2-c^2\right|}{a^2}} \)
cu egalitate daca si numai daca \( A=2C \) .
Posted: Mon Oct 01, 2007 6:33 pm
by red_dog
Inegalitatea este echivalenta cu cea obtinuta prin ridicarea ambilor membri la patrat.
\( \displaystyle\frac{b^2}{a^2}+\frac{a^2}{(b+c)^2}+\frac{2b}{b+c}\geq 2+\frac{b^2}{a^2}-\frac{c^2}{a^2}\Leftrightarrow \)
\( \displaystyle\Leftrightarrow\frac{a^2}{(b+c)^2}+\frac{c^2}{a^2}\geq\frac{2c}{b+c} \)
ultima inegalitate fiind adevarata, reprezentand de fapt inegalitatea mediilor.
Egalitate se obtine pentru \( \displaystyle\frac{a}{b+c}=\frac{c}{a} \).
Folosind teorema sinusurilor, egalitatea devine
\( \displaystyle\frac{\sin A}{\sin B+\sin C}=\frac{\sin C}{\sin A} \)
\( \sin^2A-\sin^2C=\sin B\sin C \)
\( \sin(A+C)\sin(A-C)-\sin B\sin C=0 \)
\( \sin B\sin(A-C)-\sin B\sin C=0 \)
\( \sin B(\sin (A-C)-\sin C)=0 \)
Rezulta \( \sin(A-C)=\sin C\Rightarrow A-C=C\Rightarrow A=2C \)