Question

若abc∈R＋,求证c/（a+b）＋a/（b+c）+b/（c+a）≥3/2

怎么证啊

Answer 1

证明：c/(a+b)＋a/(b+c)+b/(c+a)
=[(a+b+c)/(a+b)]-1+[(a+b+c)/(b+c)]-1+[(a+b+c)/(c+a)]-1
= -3+[(a+b+c)/(a+b)]+[(a+b+c)/(b+c)]+[(a+b+c)/(c+a)]
= -3+(a+b+c)[1/(a+b)]+[1/(b+c)]+[1/(c+a)]
将(a+b+c)拆成(1/2)[(a+b)+(b+c)+(c+a)]得
= -3+(1/2)[(a+b)+(b+c)+(c+a)][1/(a+b)]+[1/(b+c)]+[1/(c+a)]
= -3+(1/2){1+[(a+b)/(b+c)]+[(a+b)/(c+a)]+1+[(b+c)/(a+b)]+[(b+c)/(c+a)]+1+[(c+a)/(a+b)] +[(c+a) /(b+c)]}
= (-3/2)+ (1/2){[(a+b)/(b+c)]+[(b+c)/(a+b)]+ [(a+b)/(c+a)]+[(c+a)/(a+b)]+ [(b+c)/(a+b)] +[(a+b)/(b+c)]}
每两组结合，运用均值不等式得
≥(-3/2)+ (1/2)[2+2+2]
=3/2

Answer 2

解：由柯西不等式可知，[(a+b)+(b+c)+(c+a)]×[1/(a+b)+1/(b+c)+1/(c+a)]≥(1+1+1)².====>2(a+b+c)×[1/(a+b)+1/(b+c)+1/(c+a)]≥9.====>[(a+b+c)/(a+b)]+[(a+b+c)/(b+c)]+[(a+b+c)/(c+a)]≥9/2.====>{1+[c/(a+b)]}+{1+[a/(b+c)]}+{1+[b/(c+a)]}≥9/2.====>c/(a+b)+a/(b+c)+b/(c+a)≥3/2.等号仅当a=b=c时取得。

Answer 3

证明：不妨设a≤b≤c
则 c/（a+b）＋a/（b+c）+b/（c+a）≥c/（c+c）＋a/（c+c）+b/（c+c)
=（a+b+c）/2c≥3c/2c=3/2