Ta có đẳng thức:

\(a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\)

\(A=x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\ge\frac{\left(xy+yz+zx\right)^2}{3}=\frac{1}{3}\)

\(\Rightarrow Min_A=\frac{1}{3}\)khi \(x=y=z=\frac{1}{\sqrt{3}}\)

hoặc bạn áp dụng hệ thức holder á

Ta có:

\(x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\)

Mặt khác:

\(\left(xy+yz+zx\right)^2=1\le3\left(x^2y^2+y^2z^2+z^2x^2\right)\)

\(\Rightarrow\frac{1}{3}\le\left(x^2y^2+y^2z^2+z^2x^2\right)\)

hay \(x^4+y^4+z^4\ge\frac{1}{3}\Rightarrow A\ge\frac{1}{3}\)

Vậy \(Min_A=\frac{1}{3}\)khi \(x=y=z=\frac{1}{\sqrt{3}}\)

thêm x²+y²+z²=1 nha

thêm x² + y² + z² = 1 nha

      HT nha vinh

\(P=\dfrac{1}{xyz\left(x+y+z\right)}-\dfrac{2}{xy+yz+zx}\ge\dfrac{3}{\left(xy+yz+zx\right)^2}-\dfrac{2}{xy+yz+zx}\)

\(P\ge3\left(\dfrac{1}{xy+yz+zx}-\dfrac{1}{3}\right)^2-\dfrac{1}{3}\ge-\dfrac{1}{3}\)

\(P_{min}=-\dfrac{1}{3}\) khi \(x=y=z=1\)