在三角形ABC中,求证：sin(B+C)/（sinB+sinC）+sin(C+A)/（sinC+sinA）+sin(A+B)/（sinA+sinB)大于或等于3/2
热心网友
sin(B+C)/(sinB+sinC)+sin(C+A)/(sinC+sinA)+sin(A+B)/(sinA+sinB)=sinA/(sinB+sinC)+sinB/(sinC+sinA)+sinC/(sinA+sinB)=a/(b+c)+b/(c+a)+c/(a+b) 正弦定理：a=2RsinA;b=2RsinB。。。)=[a/(b+c)+1]+[b/(c+a)+1]+[c/(a+b)+1]-3=(a+b+c)/(b+c)+(a+b+c)/(c+a)+(a+b+c)/(a+b)-3=(a+b+c)[1/(b+c)+1/(c+a)+1/(a+b)]-3=[(b+c)+(c+a)+(a+b)]/2*[1/(b+c)+1/(c+a)+1/(a+b)]-3(a+b)+(b+c)+(a+b=3[(a+b)(b+c)(c+a)]*（1/3）1/（b+c)+1/(c+a)+1/(a+b)=3/[(b+c)(c+a)(a+b)]^(1/3)此二式的两边相乘得到2(a+b+c)[1/(b+c)+1/(c+a)+1/(a+b)]=9---(a+b+c)[1/(b+c)+1/(c+a)+1/(a+b)]=9/2---a/(b+c)+b/(c+a)+c/(a+b)=9/2-3=3/2。