Đáp án:
`min f = 2^(2/3) * 3^(1/2)`
Giải thích các bước giải:
`f_(x;yz) = x/y + sqrt(y/z) + 3sqrt(z/x)`
Áp dụng bất đẳng thức `Cauchy` ta có:
`f_(x;y;z) = 6*( x/y + 1/2sqrt(y/z)+1/2sqrt(y/z) +1/3* 3sqrt(z/x)+1/3* 3sqrt(z/x)+1/3* 3sqrt(z/x))/6 ge 6*6sqrt(x/y*1/2sqrt(y/z)* 1/2sqrt(y/z) * 1/3*3sqrt(z/x)* 1/3*3sqrt(z/x)* 1/3*3sqrt(z/x))`
`= 6*6sqrt(1/(2*2*3*3*3)*x/y*y/z*z/x)`
`=2^(2/3)*3^(1/2)`
Vậy `min f = 2^(2/3)*3^(1/2) <=> x/y =1/2* sqrt(y/z) =1/3* 3sqrt(z/x)`
