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Assume that $a_1,a_2,\ldots,a_n>0$ and $a_1+a_2+\cdots+a_n<\frac{1}{2}$, and prove that $$(1+a_1)(1+a_2)\cdots(1+a_n)<2$$

I've tried Hölder's inequality (the same result can easily be derived using AM-GM). I've found out that it's sufficient to prove that $$ \left(\frac{2n+1}{2n}\right)^n<2. $$

(I've created this sign for myself to use informally while searching for a proof. Proving that one of the signs holds will prove my inequality). Does anyone see how one could prove this (if it holds, of course)?

However, there must be a way to prove the inequality by using induction at first rather than the Holder's inequality or AM-GM. Thanks.

And I'm sorry. But I'd forgotten to add that neither logs nor calculus can be used.

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  • $\begingroup$ Hint: If you want to continue your proof, show that your LHS is increasing in $n$ and has a limit of $\sqrt{e} < 2$, so the inequality is strict. $\endgroup$ – Macavity Jan 28 '14 at 18:00
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First proof. We shall use the inequality $$ \log (1+x)\le x,\,\,\, \text{for $x\ge 0$}, \tag{1} $$ which implies that $$ \sum_{k=1}^n \log(1+a_n)\le \sum_{k=1}^n a_n<\frac{1}{2}, $$ and hence $$ \prod_{k=1}^n (1+a_n)=\exp\left(\sum_{k=1}^n \log(1+a_n)\right)<\mathrm{e}^{1/2}<2, $$ as $\mathrm{e}<4$.

Second proof. Without the use of $(1)$. Instead we use Cauchy Inequality: $$ \frac{x_1+\cdots+x_n}{n}\ge \sqrt[n]{x_1\cdots x_n}. $$ Here $x_i=1+a_i$, $\sum_{i=1}^n x_i=\sum_{i=1}^n(1+ a_i)<n+\frac{1}{2}$, and thus $$ \frac{n+\frac{1}{2}}{n}>\frac{x_1+\cdots+x_n}{n}\ge \left(\prod_{i=1}^n(1+a_i)\right)^{1/n}, $$ and hence $$ \prod_{i=1}^n(1+a_i)<\Big(1+\frac{1}{2n}\Big)^n<2,\tag{2} $$ since $$ \Big(1+\frac{1}{2n}\Big)^n=\sum_{k=0}^n\binom{n}{k}\left(\frac{1}{2n}\right)^{\!k}<\sum_{k=0}^n\frac{1}{2^k}<2, $$ as $$ \binom{n}{k}\left(\frac{1}{2n}\right)^{\!k}=\frac{n(n-1)\cdots(n-k+1)}{k!n^k}\cdot\frac{1}{2^k}<\frac{1}{2^k}. $$

Notes.

  1. Inequality $(1)$ is obtained by integrating $\,f(t)=\dfrac{1}{1+t}\le 1\,$ in the interval $[0,x]$.

  2. It turns out that the optimal formulation of this question is:

If $a_1,a_2,\ldots,a_n>0$ and $a_1+a_2+\ldots+a_n<\dfrac{1}{2}$, then $(1+a_1)(1+a_2)\ldots(1+a_n)<\sqrt{\mathrm{e}}$.

The optimal value is attained for $a_1=\cdots=a_n=\dfrac{1}{2n}-\varepsilon$, for $\varepsilon\to 0$ and $n\to\infty$.

It is noteworthy that $(2)$ is true if on the right $2$ is replaced by $\sqrt{\mathrm{e}}$, since $$ \left(1+\frac{1}{2n}\right)^{n}<\sqrt{\mathrm{e}}. $$

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  • $\begingroup$ Could you try thinking of a different proof? No calculus nor logs can be used. I'm sorry that I hadn't mentioned this. $\endgroup$ – user26486 Jan 28 '14 at 17:03
  • $\begingroup$ @mathh: See update. $\endgroup$ – Yiorgos S. Smyrlis Jan 28 '14 at 17:20
  • $\begingroup$ I'm sorry for asking this, but could you tell me how I could use induction for a proof? I generally understand induction, but in this case it has to be used in a different, unusual way that I can't think of. So if we know that $$a_1+a_2+\ldots+a_n<\frac{1}{2}$$then $$(1+a_1)(1+a_2)\ldots(1+a_n)<2$$ The case for $n=1$ is obvious. Now how can I continue from here? If we know that $$a_1+a_2+\ldots+a_{n+1}<\frac{1}{2}$$then $$a_1+a_2+\ldots+a_n<\frac{1}{2}$$so $$(1+a_1)(1+a_2)\ldots(1+a_{n+1})<2(1+a_{n+1})\not\lt 2$$How could I go about solving this that way? It'd be interesting to see. $\endgroup$ – user26486 Feb 21 '14 at 20:49
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Let $s_n$ denote $a_1+\ldots+a_n$. I will prove by induction that if $s_n < 1$ then

$$\prod_{k=1}^n(1+a_k) < \frac{1}{1-s_n}.$$

Your inequality is a direct consequence of this. First note that $1-a^2 < 1$ for all $a>0$ and therefore $$1+a<\frac{1}{1-a}$$ for all $a \in (0,1)$. This proves the base case $n=1$. Now by induction

$$ \begin{eqnarray} \prod_{k=1}^{n+1}(1+a_k)&=&(1+a_{n+1})\prod_{k=1}^n(1+a_k)\\ &<&\frac{1}{1-a_{n+1}}\cdot\frac{1}{1-s_n}\\ &<&\frac{1}{1-s_n-a_{n+1}}\\ &=& \frac{1}{1-s_{n+1}}. \end{eqnarray} $$

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By Bernoulli's inequality $$\left(1 - \frac{1}{2n+1}\right)^n > 1-\frac{n}{2n+1} > \frac{1}{2}$$ for $n\geq 1$ and so $$\left(1+\frac{1}{2n}\right)^n < 2.$$

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  • $\begingroup$ Nice and simple continuation of the OP's proof... (+1) $\endgroup$ – Macavity Jan 28 '14 at 18:02
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$S_0= 1$
$S_1=\sum (a_k) < \frac{1}{2} $
$S_2=\sum (a_{k_1} a_{k_2}) < (\sum (a_k))^2 < (\frac{1}{2})^2 $
......
$S_i=\sum (a_{k_1} a_{k_2} a_{k_3} a_{k_4} ....a_{k_i} ) < (\sum (a_k))^i < (\frac{1}{2})^i $
.....
$S_n=\sum (a_{k_1} a_{k_2} a_{k_3} a_{k_4} ....a_{k_n} ) < (\sum (a_k))^n < (\frac{1}{2})^n $

$\prod_{k=1}^n (1+a_n) < \sum_{k=0}^n S_{k} < \sum_{k=0}^n (\frac{1}{2})^k < 2 $

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$\left(\frac{2n+1}{2n}\right)^n = \left(1 + \frac1{2n}\right)^n$, and the derivative wrt $n$ is $n\left(1 + \frac1{2n}\right)^{n-1}\left(1 - \frac1{2n^2}\right)$ which is positive for all $n \ge 1$. So its maximum is $\lim_{n \rightarrow \infty} \left(1 + \frac1{2n}\right)^n = \sqrt{e} < 1.65 $

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For fixed $n$ and fixed $\sum{a_i}$, the maximum product occurs when the $a_i$ are equal. This follows from the fact that $\displaystyle(1+a_i)(1+a_j)\le (1+\frac{a_i+a_j}{2})^2$. (Replace any two unequal $a_k$ with their average to increase the product.)

Furthermore, for $0<\epsilon<\frac{1}{2}-\sum{a_i}$, $$\begin{align*} \displaystyle(1+a_1)(1+a_2)\ldots(1+a_n)&<\quad(1+a_1)(1+a_2)\ldots(1+a_n)(1+\epsilon)\\&<\quad\prod_{i=1}^{n+1}\left(1+\frac{1}{2(n+1)}\right)\\ &<\quad\lim_{n\rightarrow\infty}\left(1+\frac{1}{2n}\right)^n=\sqrt{e}<2 \end{align*}$$.

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