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

Some of you may have already seen this inequality. I was the one who asked for a solution:

If $a_1,\ldots,a_n>0$ and $a_1+\cdots+a_n<\frac{1}{2}$, then $(1+a_1)\cdots(1+a_n)<2$.

Now what I never found out was how to solve the inequality using induction.. which I've decided to ask in a separate question here. Here's my failed attempt to use induction here:


So if $n=1$, then $a_1<\frac{1}{2}$ and so apparently $(1+a_1)<1\frac{1}{2}<2$. This proves the base case $n=1$.

Now let's say it's true for $k$ and prove that it's true for $k+1$ as well.

If $a_1+a_2+\ldots+a_k<\frac{1}{2}$, then $(1+a_1)(1+a_2)\ldots(1+a_k)<2$.

Now what if $a_1+a_2+\ldots+a_k+a_{k+1}<\frac{1}{2}$? Then we still have that $a_1+a_2+\ldots+a_k<\frac{1}{2}$ and hence we know that $(1+a_1)(1+a_2)\ldots(1+a_k)(1+a_{k+1})<2(1+a_{k+1})\not<2$.


Any ideas would be appreciated.

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  • $\begingroup$ you didn't see this ? math.stackexchange.com/a/655092/99325 $\endgroup$ – derivative Apr 13 '14 at 20:24
  • $\begingroup$ @derivative I didn't like how general he was. You can post this special case of his answer here if you want. $\endgroup$ – user26486 Apr 13 '14 at 20:40
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We can prove this with another method: Using AM-GM inequality.

We have $\prod_{i=1}^{n} (1 + a_i) \leq (\dfrac{n + a_1 + a_2 +...+ a_n}{n})^n < (\dfrac{n + \frac12}{n})^n = \sqrt{(1 + \frac{1}{2n})^{2n}} < \sqrt{3} < 2$. We can prove by induction that: $(1 + \frac{1}{n})^n < 3$ for all naturals $n$'s , and this is well-known.

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  • $\begingroup$ Why is the following inequality true? $$\sqrt{(1 + \frac{1}{2n})^{2n}} < \sqrt{3}$$ $\endgroup$ – user26486 Apr 13 '14 at 20:30
  • $\begingroup$ @mathh $(1+\frac{1}{n})^n\to e$, so its all subsequences converges to $e$. $\endgroup$ – derivative Apr 13 '14 at 20:49

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