(用 "累次求极值法")若x,y,z ∈R+.求证:『xyz/「(1+5x)(4x+3y)(5y+6z)(z+18)」』≤「1/5120」.
问题描述：
(用 "累次求极值法")
若x,y,z ∈R+.
求证:『xyz/「(1+5x)(4x+3y)(5y+6z)(z+18)」』≤「1/5120」.
答
对于x，max(x/(1+5x)(4x+3y))时x=sqrt(3/20*y)
所以max(x/(1+5x)(4x+3y))=0.25(1/(1+5x')^2)(x'=sqrt(3/20*y)
)
同理max(z/(5y+6z)(z+18))时z=sqrt(15y)
所以max(z/(5y+6z)(z+18))=3(1/(18+z')^2)(z'=sqrt(15y)=10x')
不等式左端≤0.75y(1/(18+z')^2)(1/(1+0.5z')^2)
=0.05(15y)/((18+z')(1+0.5z'))^2=0.05(z'/(18+z')(1+0.5z'))^2
max(z'/(18+z')(1+0.5z'))时z'=6
不等式左端≤1/5120
ps:min(x/(a+bx)(c+dx))时bd*x^2=ac
sqrt()为根号
答
1+5x≥2√(5x)4x+3y≥2√(12xy)5y+6z≥2√(30yz)z+18≥2√(18z)所以:(1+5x)(4x+3y)(5y+6z)(z+18)≥16√[(5x)(12xy)(30yz)(18z)](1+5x)(4x+3y)(5y+6z)(z+18)≥2880xyzxyz/「(1+5x)(4x+3y)(5y+6z)(z+18)」』≤1/2880不...