Question

证明,在三角形ABC中,sinA/2sinB/2sinC/2<=1/8.

Answer 1

法1（sinA/2)^2+(sinB/2)^2+(sinC/2)^2+2sinA/2sinB/2sinC/2
=sinA/耐正2（（sinA/2)+2sinB/2sinC/2）+(sinB/2)^2+(sinC/2)^2
=sinA/2（（cos（B+C）/2)+2sinB/2sinC/2）+(sinB/2)^2+(sinC/2)^2
=sinA/2*cos（B+C）/2+(1-cosB)/2+(1-cosC)/2(积化和差)
=cos（B+C）/2*cos（B+C）/2+1-cos（握亩判B+C）/2*cos（B+C）/2
=1
所以（sinA/2)^2+(sinB/2)^2+(sinC/2)^2最小2sinA/2sinB/2sinC/2最大
（sinA/2)^2+(sinB/2)^2+(sinC/2)^2>=3(3次根号(（sinA/2sinB/2sinC/2）^2))
此时sinA/2=sinB/2=sinC/2
取段改等号A=B=C=60最大sinA/2sinB/2sinC/2=1/8