Ta có: \(\left(x-1\right)^3=x^3-3x^2+3x-1\)
\(=x\left(x^2-3x+3\right)-1=x\left(x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}x-1\ge\dfrac{3}{4}x-1\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\left(y-1\right)^3\ge\dfrac{3}{4}y-1;\left(z-1\right)^3\ge\dfrac{3}{4}z-1\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge\dfrac{3}{4}\left(x+y+z\right)-3=\dfrac{3}{4}\cdot3-3=-\dfrac{3}{4}\)