Ta có: $\begin{cases}x^2+y^2\geq 2xy\\y^2+z^2\geq 2yz\\ z^2+x^2\geq 2zx\end{cases}\Rightarrow x^2+y^2+z^2\geq xy+yz+zx$
$\Leftrightarrow x^2+y^2+z^2+2(xy+yz+zx)\geq 3(xy+yz+zx)\Leftrightarrow (x+y+z)^2\geq 3(xy+yz+zx).$
Thay $x=ab;y=bc;z=ca, $ ta thu được:
$(ab+bc+ca)^2\geq 3abc(a+b+c)$
Dấu $"="$ xảy ra $\Leftrightarrow a=b=c.$