Cho `x, y, z` là các số thực dương thỏa mãn điều kiện `x+y+z=1.` Chứng minh rằng `(yz)/(x^2+xyz)+(zx)/(y^2+xyz)+(xy)/(z^2+xyz)>=1/(4x)+1/(4y)+1/(4z)`

$\dfrac{yz}{x^2+xyz}=\dfrac{yz}{x(x+yz)}=\dfrac{x+yz-x}{x(x+yz)}=\dfrac{1}{x}-\dfrac{1}{x+yz}=\dfrac{1}{x}-\dfrac{1}{x^2+xy+xz+yz}=\dfrac{1}{x}-\dfrac{1}{(x+y)(x+z)}\\\to \dfrac{yz}{x^2+xyz}+\dfrac{xz}{y^2+xyz}+\dfrac{xy}{z^2+xyz}\ge \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-\dfrac{2(x+y+z)}{(x+y)(y+z)(x+z)}=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-\dfrac{2}{(x+y)(y+z)(x+z)}\\\text{Áp dụng BĐT 8/9 ta được:}\\(x+y)(y+z)(x+z)\ge \dfrac{8}{9}(x+y+z)(xy+yz+xz)=\dfrac{8}{9}(xy+yz+xz)\\\text{Áp dụng BĐT Cauchy ta được:}\\x^2y^2+y^2z^2\ge 2xy^2z, y^2z^2+x^2z^2\ge 2xyz^2, x^2y^2+x^2z^2\ge 2x^2yz\\\to x^2y^2+y^2z^2+x^2z^2\ge xyz(x+y+z)\\\to (xy+yz+xz)^2\ge 3xyz(x+y+z)=3xyz\\\to (x+y)(y+z)(x+z)\ge \dfrac{8}{9} . \dfrac{3xyz}{xy+yz+xz}=\dfrac{8xyz}{3(xy+yz+xz)}\\\to \dfrac{yz}{x^2+xyz}+\dfrac{xz}{y^2+xyz}+\dfrac{xy}{z^2+xyz}\ge \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} -\dfrac{3(xy+yz+xz)}{4xyz}=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-\dfrac{3}{4x}-\dfrac{3}{4y}-\dfrac{3}{4z}=\dfrac{1}{4x}+\dfrac{1}{4y}+\dfrac{1}{4z}\\\text{Dấu "=" xảy ra khi:}\\x=y=z=\dfrac{1}{3}$