已知正实数A B C满足1/A+2/B+3/C=1,求证A+B/2+C/3≥9
问题描述:
已知正实数A B C满足1/A+2/B+3/C=1,求证A+B/2+C/3≥9
答
(A+B/2+C/3)相乘(1/A+2/B+3/C)所得式用均值
答
柯西不等式知道吧?
A+B/2+C/3=(A+B/2+C/3)*1=(A+B/2+C/3)*(1/A+2/B+3/C)
>=(1+1+1)^2=9
当且仅当A=3,B=6,C=9时等号成立
A*(1/A),B/2*(2/B),C/3*(3/C)应该有根号,不会打,就省略啦。
答
A+B/2+C/3=(A+B/2+C/3)*1=(A+B/2+C/3)*(1/A+2/B+3/C)
>=(A*(1/A)+B/2*(2/B)+C/3*(3/C))^2=9
当且仅当A=3,B=6,C=9时等号成立