Sa se arate ca in orice triunghi avem:
\( \frac{m_{a}^{2}}{1+\cos A}+\frac{m_{b}^{2}}{1+\cos B}+\frac{m_{c}^{2}}{1+\cos C}\ge9Rr. \)
Gazeta Matematica, nr.2/2009
Inegalitate in triunghi
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Mateescu Constantin
- Newton
- Posts: 307
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Voi arata o imbunatatire a inegalitatii propuse : \( \overline{\underline{\left\|\ \sum\ \frac{m_a^2}{1+\cos A}\ \ge\ \frac{3p^2-r(4R+r)}{4}\ \ge\ r(11R-4r)\ \ge\ 9Rr\ \right\|}} \).
Demonstratie:
Voi folosi o inegalitate cunoscuta intr-un triunghi: \( \overline{\underline{\left\|\ m_a\ \ge\ \frac{b+c}{2}\ \cdot\ \cos\frac A2\ \right\|}}\ (\ast) \). Intr-adevar,
\( m_a=\frac{\sqrt{2(b^2+c^2)-a^2}}{2}=\frac{\sqrt{(b+c)^2-2bc+2bc\cos A}}{2}=\frac{\sqrt{(b+c)^2-2bc(1-\cos A)}}{2}\ \ge\ \frac{\sqrt{(b+c)^2-\frac{(b+c)^2}{2}(1-\cos A)}}{2} \)
\( \Longleftrightarrow\ m_a\ \ge\ \frac{b+c}{2}\sqrt{1-\frac{1-\cos A}{2}}=\frac{b+c}{2}\sqrt{\frac{1+\cos A}{2}}=\frac{b+c}{2}\ \cdot\ \cos\frac A2\ ,\ \mbox{\normal O.K.} \)
Prin urmare, \( \sum\ \frac{m_a^2}{1+\cos A}=\sum\ \frac{m_a^2}{2\cos^2\frac A2}\ \ge^{(\ast)}\ \sum\ \frac{\frac{(b+c)^2}{4}\cos^2\frac A2}{2\cos^2\frac A2}=\sum\ \frac{(b+c)^2}{8}=\frac{a^2+b^2+c^2+ab+bc+ca}{4} \)
\( \Longleftrightarrow\ \sum\ \frac{m_a^2}{1+\cos A}\ \ge\ \frac{2p^2-8Rr-2r^2+p^2+r^2+4Rr}{4}=\frac{3p^2-r(4R+r)}{4} \). Folosind inegalitatea Gerretsen
i.e. \( p^2\ge 16Rr-5r^2 \) obtinem in continuare : \( \frac{3p^2-4Rr-r^2}{4}\ \ge\ \frac{3(16Rr-5r^2)-4Rr-r^2}{4}=r(11R-4r) \) .
In fine, ultima inegalitate din dreapta este echivalenta cu inegalitatea lui Euler : \( R\ge 2r \) .
Demonstratie:
Voi folosi o inegalitate cunoscuta intr-un triunghi: \( \overline{\underline{\left\|\ m_a\ \ge\ \frac{b+c}{2}\ \cdot\ \cos\frac A2\ \right\|}}\ (\ast) \). Intr-adevar,
\( m_a=\frac{\sqrt{2(b^2+c^2)-a^2}}{2}=\frac{\sqrt{(b+c)^2-2bc+2bc\cos A}}{2}=\frac{\sqrt{(b+c)^2-2bc(1-\cos A)}}{2}\ \ge\ \frac{\sqrt{(b+c)^2-\frac{(b+c)^2}{2}(1-\cos A)}}{2} \)
\( \Longleftrightarrow\ m_a\ \ge\ \frac{b+c}{2}\sqrt{1-\frac{1-\cos A}{2}}=\frac{b+c}{2}\sqrt{\frac{1+\cos A}{2}}=\frac{b+c}{2}\ \cdot\ \cos\frac A2\ ,\ \mbox{\normal O.K.} \)
Prin urmare, \( \sum\ \frac{m_a^2}{1+\cos A}=\sum\ \frac{m_a^2}{2\cos^2\frac A2}\ \ge^{(\ast)}\ \sum\ \frac{\frac{(b+c)^2}{4}\cos^2\frac A2}{2\cos^2\frac A2}=\sum\ \frac{(b+c)^2}{8}=\frac{a^2+b^2+c^2+ab+bc+ca}{4} \)
\( \Longleftrightarrow\ \sum\ \frac{m_a^2}{1+\cos A}\ \ge\ \frac{2p^2-8Rr-2r^2+p^2+r^2+4Rr}{4}=\frac{3p^2-r(4R+r)}{4} \). Folosind inegalitatea Gerretsen
i.e. \( p^2\ge 16Rr-5r^2 \) obtinem in continuare : \( \frac{3p^2-4Rr-r^2}{4}\ \ge\ \frac{3(16Rr-5r^2)-4Rr-r^2}{4}=r(11R-4r) \) .
In fine, ultima inegalitate din dreapta este echivalenta cu inegalitatea lui Euler : \( R\ge 2r \) .