Xét : 1/x^2+x + x/2 + x+1/4 = 1/x.(x+1) +x/2 + x+1/4 >= 3\(\sqrt[3]{\frac{1}{x.\left(x+1\right)}.\frac{x}{2}.\frac{x+1}{4}}\) = 3/2
=> 1/x^2+x >= 3/2 - x/2 - x+1/4 = 3/2 - (3x+1)/4
Tương tự : 1/y^2+y >= 3/2 - (3y+1)/4 ; 1/z^2+z >= 3/2 - (3z+1)/4
=> M >= 9/2 - (3x+3y+3z+3)/4 = 9/2 - (3.3+3)/4 = 9/2 - 3 = 3/2
Dấu "=" xảy ra <=> x=y=z=1
Vậy GTNN của M = 3/2 <=> x=y=z=1
Tk mk nha