a,b,c∈(0,+∞,a+b+c=3,求证:a/(3-a)+b/(3-b)+c/(3-c)≥3/2=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 这步是怎么出来的?
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/25 17:47:31
![a,b,c∈(0,+∞,a+b+c=3,求证:a/(3-a)+b/(3-b)+c/(3-c)≥3/2=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 这步是怎么出来的?](/uploads/image/z/3942144-0-4.jpg?t=a%2Cb%2Cc%E2%88%88%EF%BC%880%2C%2B%E2%88%9E%2Ca%2Bb%2Bc%3D3%2C%E6%B1%82%E8%AF%81%EF%BC%9Aa%2F%283-a%29%2Bb%2F%283-b%29%2Bc%2F%283-c%29%E2%89%A53%2F2%3D3%5B1%2F%283-a%29%2B1%2F%283-b%29%2B1%2F%283-c%29%5D-3%EF%BC%88%E5%88%A9%E7%94%A8%E6%9F%AF%E8%A5%BF%E4%B8%8D%E7%AD%89%E5%BC%8F%EF%BC%89%E2%89%A53%C3%97%5B%EF%BC%881%2B1%2B1%EF%BC%89%5E2%2F%EF%BC%883-a%EF%BC%8B3-b%EF%BC%8B3-c%EF%BC%89%5D-3+%E8%BF%99%E6%AD%A5%E6%98%AF%E6%80%8E%E4%B9%88%E5%87%BA%E6%9D%A5%E7%9A%84%EF%BC%9F)
a,b,c∈(0,+∞,a+b+c=3,求证:a/(3-a)+b/(3-b)+c/(3-c)≥3/2=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 这步是怎么出来的?
a,b,c∈(0,+∞,a+b+c=3,求证:a/(3-a)+b/(3-b)+c/(3-c)≥3/2
=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
这步是怎么出来的?
a,b,c∈(0,+∞,a+b+c=3,求证:a/(3-a)+b/(3-b)+c/(3-c)≥3/2=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 这步是怎么出来的?
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