已知a,b,c是三角形ABC的三边,若a,b,c的倒数成等差数列,求证角B为锐角

2/b=1/a+1/c2/sinB=1/sinA+1/sinC2/sinB=(sinA+sinC)/sinAsinC所以sinAsinC=-(1/2)[cos(A+C)-cos(A-C)]＞0cos(A+C)-cos(A-C)＜0-cosB-cos(A-C)＜0cosB＞cos(C-A)∵1/c＞1/aa＞c∠A＞∠CC-A＜0cos(C-A)＞0∴cosB＞0B为锐角