三角形abc中,已知 π/3≤∠B≤π/2,求证a+c≤2b

π/3≤B≤π/2,,√3<=2sinB<=2,
π/4≤(A+C)/2≤π/3,√2<=2sin[(A+C)/2]<=√3
cos[(A-C)/2]<=1
sinA+sinc
=2sin[(A+C)/2]·cos[(A-C)/2]
<=√3*1
=√3
所以：sinA+sinC<=2sinB
等同于：a+c<=2b

由正弦定理可得 a+c≤2b 等价于sinA+sinC≤2sinB
左边和差化积 sinA+sinC=2sin[(A+C)/2]·cos[(A-C)/2]
A+C≤2B
若B≤π/2 则sin[(A+C)/2]≤sinB
又cos[(A-C)/2]≤1 所以成立
若π/2≤B A+C≤π/2
所以sin[(A+C)/2]≤sin(A+C)=sinB
又cos[(A-C)/2]≤1
所以成立
所以a+c≤2b