<s>[INST] 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. 
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}>\sqrt{3}$$
 [/INST] In general, for $n \ge 3$,
$$\sum_{k=1}^{n} \frac{1}{\sqrt{k}} \gt 2(\sqrt{n+1}-1) \gt \sqrt{n}$$
by using
$$\frac{1}{\sqrt{n}} = \frac{2}{2\sqrt{n}} \gt \frac{2}{\sqrt{n+1} + \sqrt{n}} = 2(\sqrt{n+1} - \sqrt{n})$$
and having a telescopic sum.
For $n \ge 3$ we also have that
$$ 2\sqrt{n+1} - 2 = \sqrt{n+1} + (\sqrt{n+1} -2) \gt \sqrt{n} $$
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