O inegalitate in triunghi

[nica]

Sa se arate ca in orice triunghi are loc inegalitatea: \( R+r\geq\sqrt[3]{r_ar_br_c} \) , notatiile sunt cele cunoscute.

[opincariumihai]

Din Gerretsen si apoi Euler obtin succesiv \( p^2\leq4R^2+4Rr+3r^2\leq5R^2+3Rr+r^2 \) de unde \( p^2r\leq5R^2r+3Rr^2+r^3\leq R^3+3R^2r+3Rr^2+r^3=(R+r)^3 \)

[Mateescu Constantin]

Din relatiile \( r_a+r_b+r_c=4R+r \) si \( \frac{1}{r}=\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c} \) inegalitatea devine

\( r_a+r_b+r_c+\frac{3}{\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c}}\ \ge\ 4\sqrt[3]{r_ar_br_c} \)

\( \Longleftrightarrow\ 3m_a+m_h\ \ge\ 4m_g \)
Avem \( m_a^2\cdot m_h\ \ge\ m_g^3 \), deoarece se reduce la \( \sum(a-b)^2\ge 0 \) .
Atunci \( 3m_a+m_h\ \ge\ 4\sqrt[4]{m_a^3\cdot m_h}=4\sqrt[4]{m_a(m_a^2\cdot m_h)}\ \ge\ 4\sqrt[4]{m_a\cdot m_g^3}\ \ge\ 4\sqrt[4]{m_g^4}=4\sqrt[3]{abc}=4m_g \)