Question

锐角三角形abc,对应边a、b、c.2a=b+c,求tan a/tanb+tana/tanc最大

Answer 1

2a=b+c,
∴b^2+c^2>=(b+c)^2/2=2a^2,当b=c时取等号,
tanA/tanB+tanA/tanC
=tanA(tanB+tanC)/(tanBtanC)
=tanAsin(B+C)/(sinBsinnC)
=(sinA)^2/(cosAsinBsinC)
=a^2/(bccosA)
=2a^2/(b^2+c^2-a^2)