Question

已知a+b+c=1,求a/(b+c)+b/(a+c)+c/(a+b)的最小值.用换元法怎么解啊

Answer 1

【注：题目还要加个条件,a,b,c>0】因a+b+c=1,故a/(b+c)=[1-(b+c)]/(b+c)=[1/(b+c)]-1.同理,b/(c+a)=[1/(c+a)]-1,c/(a+b)=[1/(a+b)]-1.设原式为S,则S+3=1/(b+c)+1/(c+a)+1/(a+b).====>2(S+3)=2[1/(b+c)+1/(c+a)+1/(a+b)]=[(b+c)+(c+a)+(a+b)]·[1/(b+c)+1/(c+a)+1/(a+b)]≥(1+1+1)²=9.(此处用的柯西不等式)===>2(S+3)≥9.===>S≥3/2.等号仅当a=b=c=1/3时取得.故Smin=3/2.即原式的最小值为3/2.