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Intr-un triunghi ABC au loc inegalitatile:

a) \( \frac{1}{2}<\frac{m_a+m_b}{a+b}<\frac{3}{2} \)

b) \( \frac{3}{10}<\frac{am_a+bm_b}{a^2+b^2}<\frac{3}{2} \)

a) Folosind formula medianei, prima parte a inegalitatii este echivalenta cu :

\( \frac{1}{2}<\frac{\frac{1}{2}(\sqrt{2(b^2+c^2)-a^2}+\sqrt{2(a^2+c^2)-b^2})}{a+b}\Leftrightarrow a+b<\sqrt{2(b^2+c^2)-a^2}+\sqrt{2(a^2+c^2)-b^2}\Leftrightarrow \)
\( 2(a+b)^2(a-b+c)(b+c-a)>0 \), evident adevarata.

Analog se face si cea de-a doua parte a inegalitatii.