\(Q=\dfrac{x^3}{y+z}+\dfrac{y^3}{x+z}+\dfrac{z^3}{x+y}\)
\(Q=\dfrac{x^4}{xy+xz}+\dfrac{y^4}{xy+zy}+\dfrac{z^4}{xz+yz}\)
\(Q\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{xy+xz+xy+zy+xz+yz}=\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(xy+yz+xz\right)}\)(svac-xo)
Lại có:\(x^2+y^2+z^2\ge xy+yz+zx\)(tự cm)
\(\Rightarrow Q\ge\dfrac{x^2+y^2+z^2}{2}\)
Mặt khác:\(3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\ge36\)(tự cm)
\(\Rightarrow x^2+y^2+z^2\ge12\)
\(\Rightarrow Q\ge\dfrac{12}{2}=6\)
Vậy MINQ=6<=>x=y=z=2
Ta có: \((\dfrac{x^3}{y+z}+\dfrac{y+z}{x})+\left(\dfrac{y^3}{x+z}+\dfrac{x+z}{y}\right)+\left(\dfrac{z^3}{x+y}+\dfrac{x+y}{z}\right)\ge2\sqrt{\dfrac{x^3\left(y+z\right)}{\left(y+z\right)x}}+2\sqrt{\dfrac{y^3\left(x+z\right)}{\left(x+z\right)y}}+2\sqrt{\dfrac{z^3\left(x+y\right)}{\left(x+y\right)z}}=2\sqrt{x^2}+2\sqrt{y^2}+2\sqrt{z^2}=2\left(x+y+z\right)\ge2.6=12\)
(Bất đẳng thức cauchy)
mà \(\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}=\dfrac{y}{x}+\dfrac{z}{x}+\dfrac{x}{y}+\dfrac{z}{y}+\dfrac{x}{z}+\dfrac{y}{z} \)
\(=\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\ge2\sqrt{\dfrac{yx}{xy}}+2\sqrt{\dfrac{zx}{xz}}+2\sqrt{\dfrac{zy}{yz}}=2+2+2=6\) (Bất đẳng thức cauchy)
\(\Rightarrow P\ge12-6=6\)
Dấu "=" xảy ra \(\Leftrightarrow\)x = y = z = 2
Vậy GTNN của P = 6 \(\Leftrightarrow\)x = y = z = 2
