ta có $\sqrt{x(y+2z)} = \frac{\sqrt{3}}{3} \sqrt{3x(y+2z)} \leq \frac{\sqrt{3}}{6} (3x+y+2z)$$\Rightarrow \frac{1}{\sqrt{x(y+2z)}}\geq \frac{1}{\frac{\sqrt{3}}{6}(3x+y+2z)}$ do đó $B= \frac{1}{\frac{\sqrt{3}}{6}(3x+y+2z)}+\frac{1}{\frac{\sqrt{3}}{6}(x+3y+2z)} +\frac{1}{\frac{\sqrt{3}}{6}(x+2y+3z)}\geq \frac{9}{\frac{\sqrt{3}}{6}(6(x+y+z)} =3$dấu "="$\Leftrightarrow x=y=z=\frac{\sqrt{3}}{3}$

ta có $\sqrt{x(y+2z)} = \frac{\sqrt{3}}{3} \sqrt{3x(y+2z)} \leq \frac{\sqrt{3}}{6} (3x+y+2z)$$\Rightarrow \frac{1}{\sqrt{x(y+2z)}}\geq \frac{1}{\frac{\sqrt{3}}{6}(3x+y+2z)}$ do đó $B\geq \frac{1}{\frac{\sqrt{3}}{6}(3x+y+2z)}+\frac{1}{\frac{\sqrt{3}}{6}(2x+3y+z)} +\frac{1}{\frac{\sqrt{3}}{6}(x+2y+3z)}\geq \frac{9}{\frac{\sqrt{3}}{6}(6(x+y+z)} =3$dấu "="$\Leftrightarrow x=y=z=\frac{\sqrt{3}}{3}$

ta xét $k\in N ,k\geq2$có : $(1+\frac{1}{k - 1} - \frac{1}{k})^{2} = 1+ \frac{1}{(k-1)^{2}}+ \frac{1}{k^{2}}+\frac{2}{k -1}- \frac{2}{k(k-1)}- \frac{2}{k}$ $= 1+\frac{1}{(k-1)^{2}}+ \frac{1}{k^{2}}+ \frac{2}{k- 1} -\frac{2}{k -1}+ \frac{2}{k} -\frac{2}{k}= 1+\frac{1}{(k-1)^{2}}+ \frac{1}{k^{2}}$ $\Rightarrow \sqrt{\frac{1}{1^{2}} +\frac{1}{(k-1)^{2}} +\frac{1}{k^{2}}} = 1+ \frac{1}{k -1}- \frac{1}{k}$ khi đó S= $1 +\frac{1}{1}- \frac{1}{2}+1+ \frac{1}{2} -\frac{1}{3}+...+1+ \frac{1}{99} -\frac{1}{100}$ =99- $\frac{1}{100}$

ta xét $k\in N ,k\geq2$có : $(1+\frac{1}{k - 1} - \frac{1}{k})^{2} = 1+ \frac{1}{(k-1)^{2}}+ \frac{1}{k^{2}}+\frac{2}{k -1}- \frac{2}{k(k-1)}- \frac{2}{k}$ $= 1+\frac{1}{(k-1)^{2}}+ \frac{1}{k^{2}}+ \frac{2}{k- 1} -\frac{2}{k -1}+ \frac{2}{k} -\frac{2}{k}= 1+\frac{1}{(k-1)^{2}}+ \frac{1}{k^{2}}$ $\Rightarrow \sqrt{\frac{1}{1^{2}} +\frac{1}{(k-1)^{2}} +\frac{1}{k^{2}}} = 1+ \frac{1}{k -1}- \frac{1}{k}$ khi đó S= $1 +\frac{1}{1}- \frac{1}{2}+1+ \frac{1}{2} -\frac{1}{3}+...+1+ \frac{1}{99} -\frac{1}{100}$ =100- $\frac{1}{100}$