Biểu thức này chỉ có min khi x;y;z là số thực dương
Đặt \(A=\dfrac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}+\dfrac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}+\dfrac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}\)
\(B=\dfrac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\dfrac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\dfrac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)
\(\Rightarrow A-B=\dfrac{x^4-y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\dfrac{y^4-z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\dfrac{z^4-x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)
\(=\dfrac{x^2-y^2}{x+y}+\dfrac{y^2-z^2}{y+z}+\dfrac{z^2-x^2}{z+x}=x-y+y-z+z-x=0\)
\(\Rightarrow A=B\)
\(\Rightarrow2A=A+B=\dfrac{x^4+y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\dfrac{y^4+z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\dfrac{z^4+x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)
\(\ge\dfrac{\left(x^2+y^2\right)^2}{2\left(x^2+y^2\right)\left(x+y\right)}+\dfrac{\left(y^2+z^2\right)^2}{2\left(y^2+z^2\right)\left(y+z\right)}=\dfrac{\left(z^2+x^2\right)^2}{2\left(z^2+x^2\right)\left(z+x\right)}\)
\(=\dfrac{x^2+y^2}{2\left(x+y\right)}+\dfrac{y^2+z^2}{2\left(y+z\right)}+\dfrac{z^2+x^2}{2\left(z+x\right)}\ge\dfrac{\left(x+y\right)^2}{4\left(x+y\right)}+\dfrac{\left(y+z\right)^2}{4\left(y+z\right)}+\dfrac{\left(z+x\right)^2}{4\left(z+x\right)}\)
\(=\dfrac{1}{2}\left(x+y+z\right)=\dfrac{1}{2}\)
\(\Rightarrow A\ge\dfrac{1}{4}\)
\(A_{min}=\dfrac{1}{4}\) khi \(x=y=z=\dfrac{1}{3}\)
