Irrational number inequality : $1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}>\sqrt{3}$

Question

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.

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

Answer 1

Hint: $\sqrt{3}+\sqrt{\frac32}+1 > 3$. Why is this true? How does it help?

Answer 2

Hint: Square $1+\frac1{\sqrt2}$ and $\sqrt3-\frac1{\sqrt3}$.

Answer 3

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

$$ 2\sqrt{n+1} - 2 = \sqrt{n+1} + (\sqrt{n+1} -2) \gt \sqrt{n} $$

Answer 4

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!!!

Answer 5

$2 \lt 3 \lt 4$ 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 $$1+\frac1{\sqrt{2}}+\frac1{\ \sqrt{3}} \gt 1+\frac7{10}+\frac47=\frac{159}{70} =\frac{636}{280}\gt \frac{630}{280}=\frac{9}{4} \gt \sqrt{5}.$$

Answer 6

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

$${\sqrt6+\sqrt3+\sqrt2\over3}\ge\sqrt[3]{\sqrt{36}}=\sqrt[6]{36}\gt\sqrt[6]8=\sqrt2$$

Answer 7

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}