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. 
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}>\sqrt{3}$$
 A: Hint: $\sqrt{3}+\sqrt{\frac32}+1 > 3$. Why is this true? How does it help?
A: Hint: Square $1+\frac1{\sqrt2}$ and $\sqrt3-\frac1{\sqrt3}$.
A: 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!!!
A: 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
$$ 2\sqrt{n+1} - 2 = \sqrt{n+1} + (\sqrt{n+1} -2) \gt \sqrt{n} $$
A: $2 \lt 3 \lt 4$ 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  $$1+\frac1{\sqrt{2}}+\frac1{\
\sqrt{3}} \gt 1+\frac7{10}+\frac47=\frac{159}{70} =\frac{636}{280}\gt \frac{630}{280}=\frac{9}{4} \gt \sqrt{5}.$$
A: 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
$${\sqrt6+\sqrt3+\sqrt2\over3}\ge\sqrt[3]{\sqrt{36}}=\sqrt[6]{36}\gt\sqrt[6]8=\sqrt2$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large%
1 + {1 \over \root{2}} + {1 \over \root{3}}}
=\pars{1 - {\root{3} \over 3}}+ \pars{{1 \over \root{2}} - {\root{3} \over 3}}
+ \pars{{1 \over \root{3}} + 2\,{\root{3} \over 3}}
\\[5mm]&={3 - \root{3} \over 3}+ {3 - \root{6} \over 3\root{2}} + \dsc{\root{3}}
=\ \overbrace{{2 \over 3 + \root{3}}}^{\ds{>\ \dsc{0}}}\ +\
\overbrace{{1 \over \root{2}\pars{3 + \root{6}}}}
^{\ds{>\ \dsc{0}}}\ +\  \dsc{\root{3}}\ \color{#66f}{\Large>\root{3}}
\end{align}
