设a,b,c大于0,求证c/(a+b)+a/(b+c)+b/(c+a)大于等于3/2.

左边
＝(a+b+c)/(a+b)+(a+b+c)/(b+c)+(a+b+c)/(c+a)-3
＝0.5×(a+b+b+c+c+a)*[1/(a+b)+1/(b+c)+1/(c+a)]-3
≥0.5×{3×[(a+b)(b+c)(c+a)]^1/3}×{3×[1/(a+b)×1/(b+c)×1/(c+a)]^1/3}-3
＝0.5×3×3-3
＝3/2
所以c/(a+b)+a/(b+c)+b/(c+a)≥3/2