a,b,c∈（0,+∞,a+b+c=3,求证：a/(3-a)+b/(3-b)+c/(3-c)≥3/2

证明：∵a,b,c∈（0,+∞）∴a/(3-a)+b/(3-b)+c/(3-c)=｛[a/(3-a)+1]＋[b/(3-b)+1]＋[c/(3-c)＋1]｝-3=[3/(3-a)+3/(3-b)+3/(3-c)]-3=3[1/(3-a)+1/(3-b)+1/(3-c)]-3（利用柯西不等式）≥3×[（1+1+1）^2/（3-a＋3-b＋3-c）]-3 =3×｛9/[9-（a+b+c)]｝-3=3×3/2-3=3/2当且仅当a=b=c=1时上式等号成立∴a/(3-a)+b/(3-b)+c/(3-c)≥3/2