Ta có: \(\left(x+6\right)^2+\left|y-\frac{1}{2}\right|+\left|x+y+z\right|\ge0\forall x;y;z\in R\)
mà theo đề bài: \(\left(x+6\right)^2+\left|y-\frac{1}{2}\right|+\left|x+y+z\right|\le0\)
\(\Rightarrow\left(x+6\right)^2+\left|y-\frac{1}{2}\right|+\left|x+y+z\right|=0\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}x=-6\\y=\frac{1}{2}\\z=\frac{11}{12}\end{cases}}\)