Cho x, y, z > 0 với mọi x + y + z = 1. Chứng minh rằng [tex]\frac{1}{x^{2} + y^{2} + z^{2}} + \frac{1}{xy} + \frac{1}{yz} + \frac{1}{xz} \geq 30[/tex]
\[\begin{align}
& \frac{1}{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\ge \frac{1}{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}+\frac{9}{xy+yz+xz} \\
& =(\frac{1}{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}+\frac{1}{xy+yz+xz}+\frac{1}{xy+yz+xz})+\frac{7}{xy+yz+xz} \\
& \ge \frac{9}{{{(x+y+z)}^{2}}}+\frac{21}{{{(x+y+z)}^{2}}}(BDT:{{(x+y+z)}^{2}}\ge 3(xy+yz+xz)) \\
& =30 \\
& ''=''\Leftrightarrow x=y=z=\frac{1}{3} \\
\end{align}\]