[gools]

\[\begin{array}{rcl} &&\frac{(x+y+z)(x^{2}+y^{2}+z^{2})}{xy+yz+zx} +3 \geq 2(x+y+z) \\
&\Leftrightarrow& (x+y+z)(x^{2}+y^{2}+z^{2})+3(xy+yz+zx) \geq 2(x+y+z)(xy+yz+zx) \\
&\Leftrightarrow& (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)+3(xy+yz+zx) \geq (x+y+z)(xy+yz+zx) \\
&\Leftrightarrow& x^3+y^3+z^3-3xyz+3(xy+yz+zx) \geq (x+y+z)(xy+yz+zx) \end{array}\]
เนื่องจาก \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\) ดังนั้น \(xy+yz+zx=3xyz\)
ดังนั้น
\[\begin{array}{rcl} && x^3+y^3+z^3-3xyz+3(xy+yz+zx) \geq (x+y+z)(xy+yz+zx) \\
&\Leftrightarrow& x^3+y^3+z^3-3xyz+9xyz \geq (x+y+z)(xy+yz+zx) \\
&\Leftrightarrow& x^3+y^3+z^3+6xyz \geq x^2y+x^2z+y^2z+y^2x+z^2x+z^2y+3xyz \\
&\Leftrightarrow& x^3+y^3+z^3-(x^2y+x^2z+y^2z+y^2x+z^2x+z^2y)+3xyz \geq 0 \\
&\Leftrightarrow& x(x-y)(x-z)+y(y-z)(y-x)+z(z-x)(z-y) \geq 0
\end{array}
\]
ซึ่งเป็นจริงตาม Schur's Inequality