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Let $S=a_1+...+a_n<1$ where $a_i>0$. Prove that $1+S<(1+a_1)\cdot ... \cdot (1+a_n)<{1\over 1-S}$. I started with the right inequality but I am not sure it iss plausible (I did something invalid I suppose.)

Attempt of right inequality: By induction on $n$. For $n=1$ we get that $0<S=a_1<1$ and therefore $(1+a_1)<?{1\over1-S} $$\to$ $(1+a_1)(1-a_1)=1-a_1^{2}<1$ . Now assume the inequality holds for $n$ and show it also does for $n+1$: $(1+a_1)\cdot...\cdot (1+a_n)\cdot (1+a_{n+1})<?{1\over 1- (a_1...+a_{n+1})}$. Now let us replace $(1+a_1)\cdot...\cdot (1+a_n)$ by ${1\over 1-(a_1...+a_{n})}$ which is greater by assumption. If the inequality still holds, it will as well for $(1+a_1)\cdot...\cdot (1+a_n)$ in particular.

${1\over 1-(a_1...+a_{n})}(1+a_{n+1})<?{1\over 1- (a_1...+a_{n+1})}$. We get $(1+a_{n+1})(1- (a_1...+a_{n}))<?1-(a_1...+a_{n})$ $\Rightarrow$
$1-(a_1...+a_{n+1})+a_{n+1}-a_{n+1}(a_1...+a_{n+1})<?1-(a_1...+a_{n})$ $\Rightarrow$ $1-(a_1...+a_{n+1})-a_{n+1}(a_1...+a_{n+1})<?1-(a_1...+a_{n+1})$ $\Rightarrow$ $-a_{n+1}(a_1...+a_{n+1})<?0$ which is correct. Therefore the right inequality holds for ant natural number. As for the left one, I am still having troubles.

I would truly appreciate your observation and help.

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  • $\begingroup$ Note the rhs is $\sum_0^\infty S^k$ $\endgroup$ – Mark Jan 30 '15 at 14:18
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You only wanted to prove the left inequality.

Then just note by expanding $\prod (1+a_n)$ we can get at least $1+S$ since we can select 1 each time, and we can select any $a_k$ once and 1 all other times when computing terms of the expansion.

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  • $\begingroup$ Thank you very much. That was really helpful. $\endgroup$ – Meitar Abarbanel Jan 30 '15 at 14:28
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The expansion of the product $(1 + a_1)\cdots (1 + a_n)$ contains $1$ as a term and $a_1 + \cdots a_n = S$ as a term. Since all other terms are positive, $(1 + a_1)\cdots (1 + a_n) > 1 + S$.

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  • $\begingroup$ Thank you two very much. From some reason I thought I had to solve the whole expression... $\endgroup$ – Meitar Abarbanel Jan 30 '15 at 14:28
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Denote inequalities 1,2,3 in the order you have. Clearly 3> 1 by binomial theorem. Now log 2 and 3. Expand 3 in maclaurin series. You can do it due to properties of S. For 2 use the property of log functin and get the upper bound: $\log (1+a) <a $. Now the inequalitu is easy.

Use the same logic to show 2> 1. Again kep in mind this is posdible because all terms are positive and their sum is less tan 1.

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