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

Or, one can consider the sum $\sqrt{3} + \sqrt{3/2} + 1$; we have
$\sqrt{3} > 1, \tag{1}$
$\sqrt{\dfrac{3}{2}} > 1, \tag{2}$
$1 \ge 1, \tag{3}$
(this last inequality being particularly subtle!) and adding (1)-(3) we have
$\sqrt{3} + \sqrt{\dfrac{3}{2}} + 1 > 3; \tag{4}$
now we divide by $\sqrt{3}$, and voila!, we obtain
$1 + \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{3}} > \sqrt{3}, \tag{5}$
the long-sought result!
With a tip-o'-the-hat to Eric Stucky, tho' I saw not his answer 'til mine nearly written was! (Yodaspeak!)
Note Added 24 November 2014, 1:02 PM PST:  We can generalize:  since 
$\sqrt{\dfrac{n}{k}} > 1, \; \; 1 \le k < n, \tag{6}$
we can, again invoking (3), write
$\sum_{k = 1}^n \sqrt{\dfrac{n}{k}} > n, \tag{7}$
and now dividing by $\sqrt{n}$ yields
$\sum_1^n \dfrac{1}{\sqrt{k}} > \sqrt{n}. \tag{8}$
End of Note.
Hope this helps.  Cheers,
and as ever,
Fiat Lux!!!
