Fie un triunghi \( ABC \) cu centrul de greutate \( G \) si centrul cercului inscris \( I \).
Sa se arate ca exista inegalitatea \( \overline{\underline{\left\|\ \frac {AG}{AI}+\frac {BG}{BI}+\frac {CG}{CI}\ \ge\ \frac 13\cdot (a+b+c)\cdot\left(\frac 1a+\frac 1b+\frac 1c\right)\ \right\|}} \).
Inegalitate intr-un triunghi.
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Mateescu Constantin
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Se observa ca : \( LHS=\sum\frac{\frac 23\cdot m_a}{\frac{b+c}{a+b+c}\cdot l_a}=\frac{2(a+b+c)}{3}\ \cdot\ \sum\frac{m_a}{l_a(b+c)} \) .
Prin urmare va fi suficient sa aratam ca \( \sum\frac{m_a}{l_a(b+c)}\ \ge\ \frac 12\left\(\frac 1a+\frac 1b+\frac 1c\right\) \) .
Dar in orice triunghi este adevarata inegalitatea : \( \overline{\underline{\left\|\ m_a\ \ge\ \frac{b+c}{2}\ \cdot\ \cos\frac A2\ \right\| \) .
Intr-adevar, \( m_a=\frac{\sqrt{2(b^2+c^2)-a^2}}{2}=\frac{\sqrt{(b+c)^2-2bc+2bc\cos A}}{2}=\frac{\sqrt{(b+c)^2-2bc(1-\cos A)}}{2}\ \ge\ \frac{\sqrt{(b+c)^2-\frac{(b+c)^2}{2}(1-\cos A)}}{2} \)
\( \Longleftrightarrow\ m_a\ \ge\ \frac{b+c}{2}\sqrt{1-\frac{1-\cos A}{2}}=\frac{b+c}{2}\sqrt{\frac{1+\cos A}{2}}=\frac{b+c}{2}\ \cdot\ \cos\frac A2 \)
\( \Longrightarrow\ \frac{m_a}{l_a(b+c)}\ \ge\ \frac{\frac{b+c}{2}\cos\frac A2}{\frac{2bc}{b+c}\cos\frac A2\ \cdot\ (b+c)}=\frac{b+c}{4bc} \) . Deci \( \sum\frac{m_a}{l_a(b+c)}\ \ge\ \sum\frac{b+c}{4bc}=\frac 12\left\(\frac 1a+\frac 1b+\frac 1c\right\) \) .
Remarca.
In rezolvarea problemei esentiala a fost inegalitatea \( m_a\ \ge\ \frac{b+c}{2}\ \cdot\ \cos\frac A2 \) , "mai tare" decat \( m_a\ \ge\ \sqrt{p(p-a)} \) .
Prin urmare va fi suficient sa aratam ca \( \sum\frac{m_a}{l_a(b+c)}\ \ge\ \frac 12\left\(\frac 1a+\frac 1b+\frac 1c\right\) \) .
Dar in orice triunghi este adevarata inegalitatea : \( \overline{\underline{\left\|\ m_a\ \ge\ \frac{b+c}{2}\ \cdot\ \cos\frac A2\ \right\| \) .
Intr-adevar, \( m_a=\frac{\sqrt{2(b^2+c^2)-a^2}}{2}=\frac{\sqrt{(b+c)^2-2bc+2bc\cos A}}{2}=\frac{\sqrt{(b+c)^2-2bc(1-\cos A)}}{2}\ \ge\ \frac{\sqrt{(b+c)^2-\frac{(b+c)^2}{2}(1-\cos A)}}{2} \)
\( \Longleftrightarrow\ m_a\ \ge\ \frac{b+c}{2}\sqrt{1-\frac{1-\cos A}{2}}=\frac{b+c}{2}\sqrt{\frac{1+\cos A}{2}}=\frac{b+c}{2}\ \cdot\ \cos\frac A2 \)
\( \Longrightarrow\ \frac{m_a}{l_a(b+c)}\ \ge\ \frac{\frac{b+c}{2}\cos\frac A2}{\frac{2bc}{b+c}\cos\frac A2\ \cdot\ (b+c)}=\frac{b+c}{4bc} \) . Deci \( \sum\frac{m_a}{l_a(b+c)}\ \ge\ \sum\frac{b+c}{4bc}=\frac 12\left\(\frac 1a+\frac 1b+\frac 1c\right\) \) .
Remarca.
In rezolvarea problemei esentiala a fost inegalitatea \( m_a\ \ge\ \frac{b+c}{2}\ \cdot\ \cos\frac A2 \) , "mai tare" decat \( m_a\ \ge\ \sqrt{p(p-a)} \) .