Question

已知三角形abc三内角a,b,c成等差数列，求证：对应三边a,b,c满足1/(a+b）+1/(b+c)=

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Answer 1

证明：a+b+c=180°，2b=a+c=180°-b，则b=60°；
则由余弦定理可知：cosb=(a²+c²-b²)/(2ac)=cos60°=1/2
即(a²+c²-b²)/(2ac)=1/2
a²+c²-b²=ac
a²+c²=ac+b²
a²+c²+ab+bc=ac+b²+ab+bc
c(b+c)+a(a+b)=a(b+c)+b(b+c)=(a+b)(b+c)
[c(b+c)+a(a+b)]/[(a+b)(b+c)]=1
[c/(a+b)]+[a/(b+c)]=1
[c/(a+b)]+1+[a/(b+c)]+1=1+1+1
[c/(a+b)]+[(a+b)/(a+b)]+[a/(b+c)]+[(b+c)/(b+c)]=3
[(a+b+c)/(a+b)]+[(a+b+c)/(b+c)]=3
[1/(a+b)]+[1/(b+c)]=3/(a+b+c)

Answer 2

求证什么?等式都不成立