Question

cho 3 số thực dưỡng, y, z thỏa mản xyz=1. tìm GTLN của

A=1/(x^3 +y^3 +1) + 1/(y^3 + z^3 + 1) + 1/(z^3 + x^3 +1)

Answer 1

\(A=\frac{1}{x^3+y^3+1}+\frac{1}{y^3+z^3+1}+\frac{1}{z^3+x^3+1}\)

Ta có: 

\(x^3+y^3+xyz=\left(x+y\right)\left(x^2+y^2-xy\right)+xyz\ge xy\left(x+y+z\right)\)

Tương tự:

\(y^3+z^3+xyz\ge yz\left(x+y+z\right);\)\(z^3+x^3+xyz\ge zx\left(x+y+z\right)\)

\(\Rightarrow A\le\frac{1}{xy\left(x+y+z\right)}+\frac{1}{yz\left(x+y+z\right)}+\frac{1}{zx\left(x+y+z\right)}\)

\(\Rightarrow A\le\frac{1}{x+y+z}\cdot\frac{x+y+z}{xyz}=\frac{1}{xyz}=1\)

Dấu = khi x=y=z