sinA+sin[(C-B)]=2sin[(A+C-B)/2]cos[A+B-C)/2]
=2sin[(180°-2B)/2]cos[(180°-2C)/2]
=2sin(90°-B)cos(90°-C)
=2cosBsinC
∴tanB=[cos(C-B)]/[sinA+sin(C-B)]=[cos（C-B）]/2cosBsinC
2sinBsinC=cosBcosC+sinBsinC
cosBcosC-sinBsinC=0
cos（B+C）=0
B+C=90°
(b+c)/a=(sinb+sinc)/sina=sinb+sinc
=sinb+sin(90-b)
=√2(sin(b+45))
小于=√2
大于1不用证了吧!