Irrational number inequality : $1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}>\sqrt{3}$ it is easy and simple I know but still do not know how to show it (obviously without simply calculating the sum but manipulation on numbers is key here. 
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}>\sqrt{3}$$

The inequality $1+{1\over\sqrt2}+{1\over\sqrt3}\gt\sqrt3$ is equivalent to
$${\sqrt6+\sqrt3+\sqrt2\over3}\gt\sqrt2$$
If you accept the arithmetic-geometric mean inequality $(a+b+c)/3\ge\sqrt[3]{abc}$, then
$${\sqrt6+\sqrt3+\sqrt2\over3}\ge\sqrt[3]{\sqrt{36}}=\sqrt[6]{36}\gt\sqrt[6]8=\sqrt2$$
