Giải thích các bước giải:
$\dfrac{1}{x^2+y^2+z^2}+\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}$
$\ge\dfrac{1}{x^2+y^2+z^2}+\dfrac{9}{xy+yz+zx}$
$\ge\dfrac{1}{x^2+y^2+z^2}+\dfrac{1}{xy+yz+zx}+\dfrac{1}{xy+yz+zx}+\dfrac{7}{xy+yz+zx}$
$\ge\dfrac{9}{x^2+y^2+z^2+xy+yz+zx+xy+yz+zx}+\dfrac{7}{\dfrac{(x+y+z)^2}{3}}$
$\ge\dfrac{9}{(x+y+z)^2}+\dfrac{7}{\dfrac{(x+y+z)^2}{3}}$
$=30$
Dấu = xảy ra khi $x=y=z=\dfrac 13$
