Question

a , cho x,y,z >0 ; xyz =1 

CMR: \(\frac{x^3}{\left(1+y\right).\left(1+z\right)}\)+\(\frac{y^3}{\left(1+z\right).\left(1+x\right)}\)+\(\frac{z^3}{\left(1+x\right).\left(1+y\right)}\ge\frac{3}{4}\)

Answer 1

cha ôi rk mà cx ko bt

Answer 2

khó vcl

Answer 3

Cái này làm theo Bunhiacopski.

Answer 4

Áp dụng BĐT Cô-si cho 3 số không âm, ta có: \(\frac{x^3}{y+1.z+1}+\frac{y+1}{8}+\frac{z-1}{8}\ge3.\sqrt[3]{\frac{x^3}{64}}=\frac{3x}{4}\)

Tương tự ta có: \(\frac{y^3}{x+1.z+1}\ge\frac{3y}{4};\frac{z^3}{x+1.y+1}\ge\frac{3z}{4}\)

\(\Rightarrow\frac{x^3}{1+y.1+z}+\frac{y^3}{1+z.1+x}+\frac{z^3}{1+x.1+y}\ge\frac{1}{2}x+y+z-\frac{3}{4}\)

Mà: \(xyz=1\Rightarrow x+y+z\ge\sqrt[3]{xyz}=1\)

\(\RightarrowĐPCM\)