利用不等式求最值:求a/(b+c)+4b/(a+c)+5c/(a+b)的最小值,abc均大于0求a/(b+c)+4b/(a+c)+5c/(a+b)的最小值.abc均大于0
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![利用不等式求最值:求a/(b+c)+4b/(a+c)+5c/(a+b)的最小值,abc均大于0求a/(b+c)+4b/(a+c)+5c/(a+b)的最小值.abc均大于0](/uploads/image/z/2725521-33-1.jpg?t=%E5%88%A9%E7%94%A8%E4%B8%8D%E7%AD%89%E5%BC%8F%E6%B1%82%E6%9C%80%E5%80%BC%EF%BC%9A%E6%B1%82a%2F%EF%BC%88b%2Bc%EF%BC%89%2B4b%2F%EF%BC%88a%2Bc%EF%BC%89%2B5c%2F%EF%BC%88a%2Bb%EF%BC%89%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%2Cabc%E5%9D%87%E5%A4%A7%E4%BA%8E0%E6%B1%82a%2F%EF%BC%88b%2Bc%EF%BC%89%2B4b%2F%EF%BC%88a%2Bc%EF%BC%89%2B5c%2F%EF%BC%88a%2Bb%EF%BC%89%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC.abc%E5%9D%87%E5%A4%A7%E4%BA%8E0)
利用不等式求最值:求a/(b+c)+4b/(a+c)+5c/(a+b)的最小值,abc均大于0求a/(b+c)+4b/(a+c)+5c/(a+b)的最小值.abc均大于0
利用不等式求最值:求a/(b+c)+4b/(a+c)+5c/(a+b)的最小值,abc均大于0
求a/(b+c)+4b/(a+c)+5c/(a+b)的最小值.abc均大于0
利用不等式求最值:求a/(b+c)+4b/(a+c)+5c/(a+b)的最小值,abc均大于0求a/(b+c)+4b/(a+c)+5c/(a+b)的最小值.abc均大于0
a/(b+c)+4b/(a+c)+5c/(a+b)
=a/(b+c)+1+4b/(a+c)+4+5c/(a+b)+5-10
=(a+b+c)/(b+c)+4(a+b+c)/(a+c)+5(a+b+c)/(a+b)-10
=(a+b+c)(1/(b+c)+4/(a+c)+5/(a+b))-10
=1/2*(b+c+a+c+a+b)(1/(b+c)+4/(a+c)+5/(a+b))-10
利用柯西不等式(或者自己证明(x+y+z)(1/x+4/y+5/z)>(1+2+sqrt(5))^2,展开即可)有上式>=1/2*(1+2+sqrt(5))^2-10=3sqrt(5)-3
当且仅当a/(sqrt(5)/10-1/4)=b/(sqrt(5)/10+1/4)=c/(3/4-sqrt(5)/10)
故最小值为3sqrt(5)-3
sqrt(5)表示5开根号
解:设x=b+c,y=a+c,z=a+b,则
(x+y+z)/2=a+b+c
a=(x+y+z)/2-x,b=(x+y+z)/2-y,c=(x+y+z)/2-z
原式=[(x+y+z)/2-x]/x+4[(x+y+z)/2-y]/y+5[(x+y+z)/2-z]/z
=[(x+y+z)/2](1/x+4/y+5/z)-10
=[(x+y+z)(1/x+4/y...
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解:设x=b+c,y=a+c,z=a+b,则
(x+y+z)/2=a+b+c
a=(x+y+z)/2-x,b=(x+y+z)/2-y,c=(x+y+z)/2-z
原式=[(x+y+z)/2-x]/x+4[(x+y+z)/2-y]/y+5[(x+y+z)/2-z]/z
=[(x+y+z)/2](1/x+4/y+5/z)-10
=[(x+y+z)(1/x+4/y+5/z)-20]/2
根据柯西不等式,得
(x+y+z)(1/x+4/y+5/z)≥[(√x)(√1/√x)+(√y)(√4/√y)+(√z)(√5/√z)]²
=(1+2+√5)²=14+6√5
当且仅当(√x)/(√1/√x)=(√y)/(√4/√y)=(√z)/(√5/√z)时,取等
此时x=y /2=z /√5
因为原式=[(x+y+z)(1/x+4/y+5/z)-20]/2≥(14+6√5-20)/2=3(√5-1)
所以原式的最小值为3(√5-1)
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