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. 
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}>\sqrt{3}$$

$2 \lt 3 \lt 4$ so   $\sqrt{2} \lt \sqrt{3} \lt 2$ and 
 $$1+\frac1{\sqrt{2}}+\frac1{\
\sqrt{3}} \gt 1+\frac12+\frac12 = 2 \gt \sqrt{3}.$$
Of course that does not generalise in any nice way.
However $2 \lt \frac{100}{49}$ with $3 \lt \frac{49}{16}$ and $5 \lt \frac{81}{16}$
does yield  $$1+\frac1{\sqrt{2}}+\frac1{\
\sqrt{3}} \gt 1+\frac7{10}+\frac47=\frac{159}{70} =\frac{636}{280}\gt \frac{630}{280}=\frac{9}{4} \gt \sqrt{5}.$$
