$\displaystyle \begin{array}{{>{\displaystyle}l}} \sqrt{3} \ và\ 2\ \\ Ta\ có\ :\ 2=\sqrt{4} \ \\ \sqrt{4} >\sqrt{3} \ hay\ 2 >\sqrt{3} \ \\ \sqrt{11} +\sqrt{3} \ và\ \sqrt{2} \ \\ \left(\sqrt{11} +\sqrt{3}\right)^{2} =14+2\sqrt{33} \ \\ Ta\ có\ \sqrt{11} \ >\sqrt{2} \ \\ \sqrt{3} \ >\sqrt{2} \ \\ \rightarrow \sqrt{11} +\sqrt{3} >\sqrt{2} \ \\ \sqrt{2} +\sqrt{3} \ và\ 10\ \\ 10=7+3\ =\sqrt{49} +\sqrt{9} \ \\ Ta\ có\ :\ \sqrt{2} \ < \sqrt{9} \ \\ \sqrt{49} \ >\sqrt{3} \ \\ \rightarrow \sqrt{2} +\sqrt{3} \ < \sqrt{49} +\sqrt{9}\\ hay\ \sqrt{2} +\sqrt{3} < 10\ \\ \\ \end{array}$
