在三角形ABC中，求证：sin^2A+sin^2B+sin^2C小于等于9/4

原式=(1-cosA)/2
+(1-cosB)/2
+(1-cos^2C)
=2-cos(A+B)cos(A-B)-cos^2C
=2+cosCsoc(A-B)-cos^2C<=2+|cosC|-cos^C=-(|cosC|-1/2)^2+9/4
当cosC=1/2时(及A=B=C=60度)有最大值9/4
,
m>0
因为a+b>c,所以a+b-c>0<