Dễ dàng c/m: $(x+y+z)^2 \ge 3(xy+yz+zx) \forall x,y,z$$\Rightarrow \frac{xy+yz+zx}{x+y+z} \le \frac{x+y+z}{3}\Rightarrow \frac{xy+yz+zx}{xyz} \le \frac{x+y+z}{3}\Rightarrow \sum\frac1x \le \frac{x+y+z}{3}$
Ta có
$ VT=\sum \frac1 x+\sum \frac{\sqrt{1+x^2}}{x} $
$\le \frac{x+y+z}3+\sum \frac{\sqrt{2}. \frac{\sqrt{x^2+1}}{\sqrt{2}}}{x} \overset{Côsi}{\le} \frac{x+y+z}3+\sum\frac{2+\frac{x^2+1}{2}}{2x} $
$=\frac{x+y+z}3+\sum\frac{x^2+5}{4x}=\frac{x+y+z}3+\sum\frac{x}{4}+\frac54( \sum\frac{1}{x})$
$\le \frac{x+y+z}3+\frac{x+y+z}4+\frac{5}{4}.\frac{x+y+z}3=x+y+z=xyz=VP$
Dấu $"="$ xảy ra khi $x=y=z=\sqrt3$ @@