KöMaL 2015 B4692 The sides of an acute-angled triangle are aaa , bbb , ccc , the opposite angles are α\alphaα, β\betaβ, γ\gammaγ, and the lengths of the corresponding altitudes are mam_ama, mbm_bmb, mcm_cmc, respectively. Prove that maa+mbb+mcc≥2cosαcosβcosγ(1sin2α+1sin2β+1sin2γ)+3 \frac{m_a}{a} + \frac{m_b}{b} + \frac{m_c}{c} \ge 2\cos \alpha \cos\beta \cos \gamma \left(\frac{1}{\sin 2\alpha} + \frac{1}{\sin 2\beta} + \frac{1}{\sin 2\gamma}\right) + \sqrt{3}\,ama+bmb+cmc≥2cosαcosβcosγ(sin2α1+sin2β1+sin2γ1)+3. Official text KöMaL. Quoted with source. Solutions No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label. Add a solution Similar problems These problems have the most words in common with this statement. KöMaL 2025 B5441KöMaL 2014 B4628DE-MO 1965 Problem 6NMC 2017 Problem 2KöMaL 2012 B4427 PreviousB4691 NextB4693 Locator hu-komal/2015/b4692 · Tag V12FRBVMN3 · Edit this page on GitHub · Keys: j next, k previous