Giải thích các bước giải:
Ta có:
$P=x^2+y^2+z^2+\dfrac{1}{x+y+z}$
$\rightarrow P\ge \dfrac{(x+y+z)^2}{3}+\dfrac{1}{x+y+z}$
$\rightarrow P\ge \dfrac{17(x+y+z)^2}{54}+\dfrac{(x+y+z)^2}{54}+\dfrac{1}{2(x+y+z)}+\dfrac{1}{2(x+y+z)}$
$\rightarrow P\ge \dfrac{17(3\sqrt[3]{x.y.z})^2}{54}+3\sqrt[3]{\dfrac{(x+y+z)^2}{54}.\dfrac{1}{2(x+y+z)}.\dfrac{1}{2(x+y+z)}}$
$\rightarrow P\ge \dfrac{17(3.1)^2}{54}+\dfrac{1}{2}$
$\rightarrow P\ge \dfrac{10}{3}$
Dấu = xảy ra khi $x=y=z=1$