因为tanα + tanβ = sinα / cosα + sinβ / cosβ = (sinαcosβ+sinβcosα)/cosαcosβ = sin(α+β)/cosα cosβ
所以sin²(α+β) / (tanα + tanβ) = sin(α+β)cosα cosβ
= sin(α+β)[cos(α+β)+cos(α-β)]/2
≤|sin(α+β)|*[cos(α+β)+1]/2
= sqrt[(1-cos²(α+β))(1+cos(α+β))²] /2
= sqrt[(3-3cos(α+β)(1+cos(α+β))(1+cos(α+β))(1+cos(α+β)) /3] /2
≤sqrt[ ([(3-3cos(α+β)+(1+cos(α+β))+(1+cos(α+β))+(1+cos(α+β))]/4 )⁴ /3] /2
= sqrt[(6/4)⁴ /3] /2 = 3sqrt(3)/8
当α=β=π/6 时等号成立