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Let $x_1,\ldots, x_n$ be positive integers. Use mathematical induction to prove that

$$\left(\sum_{i=1}^n x_i\right) \cdot \left(\sum_{i=1}^n \frac{1}{x_i}\right) \ge n^2$$ for all integers $n \ge 1$.

(Given Hint: For all positive integers $a$ and $b$, $\frac{a}{b}+\frac{b}{a} \ge 2$.)

Can anyone help? Thank you.


marked as duplicate by haqnatural, PSPACEhard, Joey Zou, Winther, Parcly Taxel Sep 8 '16 at 5:02

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  • $\begingroup$ It's easy to prove this without using induction: $\left(\sum_{i=1}^n x_i\right) \bullet \left(\sum_{i=1}^n \frac{1}{x_i}\right) = \frac{n(n+1)}{2} H_n \ge n^2 \ln(n) / 2,$ which is greater than or equal to $n^2$ if $\ln n \ge 2,$ that is if $n \ge 8.$ Check it for small numbers by hand and you're done. $\endgroup$ – Mitchell Spector Sep 8 '16 at 4:43

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

This answer $\ds{\underline{is\ not}}$ based in 'Induction' as the OP requested. @RobertZ just called this point to my attention. Indeed, it's a straightforward point of view.

\begin{align} \color{#f00}{\pars{\sum_{i = 1}^{n}x_{i}}\pars{\sum_{i = 1}^{n}{1 \over xi}}} & = \sum_{i = 1}^{n}\sum_{k = 1}^{n}{x_{i} \over x_{k}} = \half\sum_{i = 1}^{n}\sum_{k = 1}^{n} \pars{{x_{i} \over x_{k}} + {x_{k} \over x_{i}}} \\[5mm] & = \half\sum_{i = 1}^{n}\sum_{k = 1}^{n} \bracks{\pars{\root{x_{i} \over x_{k}} - \root{x_{k} \over x_{i}}}^{2} + 2} \geq \half\sum_{i = 1}^{n}\sum_{k = 1}^{n}2 = \color{#f00}{n^{2}} \end{align}

  • $\begingroup$ Yours is simpler in some way, but I tried to follow OP's indications: use mathematical induction and the hint. $\endgroup$ – Robert Z Sep 8 '16 at 6:05
  • $\begingroup$ @RobertZ It's my fault. I didn't read completely the whole question. Anyway, it's another approach to add to the OP question. Thanks. $\endgroup$ – Felix Marin Sep 8 '16 at 6:16

Suppose the given inequality holds for an integer $n(\geq 1)$.

Then for $n+1$,

$$\left(\sum_{i=1}^{n+1} x_i\right)\left(\sum_{i=1}^{n+1} \frac{1}{x_i}\right)=\left(\sum_{i=1}^{n} x_i+x_{n+1}\right)\left(\sum_{i=1}^{n} \frac{1}{x_i}+\frac{1}{x_{n+1}}\right)$$

Expanding this gives us $$=\left(\sum_{i=1}^{n} x_i\right)\left(\sum_{i=1}^{n} \frac{1}{x_i}\right)+x_{n+1}\sum_{i=1}^{n} \frac{1}{x_i}+\frac{1}{x_{n+1}}\sum_{i=1}^{n} x_i+1$$

Since the given inequality holds for the integer $n$, the first term is larger than $n^2$. So,

$$\geq n^2+x_{n+1}\sum_{i=1}^{n} \frac{1}{x_i}+\frac{1}{x_{n+1}}\sum_{i=1}^{n} x_i+1$$

Now we show that $x_{n+1}\sum_{i=1}^{n} \frac{1}{x_i}+\frac{1}{x_{n+1}}\sum_{i=1}^{n} x_i$ is larger than $2n$.

This is where we use the hint. Expanding the $\sum$ gives us $$x_{n+1}\left(\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n}\right)+\frac{1}{x_{n+1}}\left(x_1+x_2+\cdots+x_n\right)$$

Rearranging the terms, we get

$$\left( \frac{x_{n+1}}{x_1}+\frac{x_1}{x_{n+1}} \right)+\left( \frac{x_{n+1}}{x_2}+\frac{x_2}{x_{n+1}} \right)+\cdots+\left( \frac{x_{n+1}}{x_n}+\frac{x_n}{x_{n+1}} \right)$$

and using the hint, this is larger than $2n$


$$\left(\sum_{i=1}^{n+1} x_i\right)\left(\sum_{i=1}^{n+1} \frac{1}{x_i}\right)\geq n^2+2n+1=(n+1)^2$$

The inequality holds for $n+1$.

  • $\begingroup$ Thank you very much...appreciate..;) $\endgroup$ – Steven Lau Zhou Sheng Sep 7 '16 at 8:52

Hint for the inductive step.

Let $A_n=\sum_{i=1}^{n} x_i$ and let $B_n=\sum_{i=1}^{n} \frac{1}{x_i}$. Then $$A_{n+1}\cdot B_{n+1}=\left (A_n+ x_{n+1}\right)\cdot \left (B_n+\frac{1}{x_{n+1}} \right)=A_nB_n+\frac{A_n}{x_{n+1}}+B_n x_{n+1}+1\\\geq n^2+1 +\frac{A_n}{x_{n+1}}+B_nx_{n+1}\stackrel{?} {\geq}(n+1)^2=n^2+2n+1.$$ So it suffices to show that $$\frac{A_n}{x_{n+1}}+B_n x_{n+1}\geq 2n.$$ Now, since $(u+v)\geq 2\sqrt{uv}$, for $u,v>0$, $$\frac{A_n}{x_{n+1}}+B_n x_{n+1}\geq 2\sqrt{A_n B_n}\geq 2n$$ and we are done.

  • $\begingroup$ After I wrote my above comment, I found some simpler way. I still keep the upvote for your answer. $\endgroup$ – Felix Marin Sep 8 '16 at 4:40

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