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.

  • $\begingroup$ Note the rhs is $\sum_0^\infty S^k$ $\endgroup$ – Mark Jan 30 '15 at 14:18

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.

  • $\begingroup$ Thank you very much. That was really helpful. $\endgroup$ – Meitar Abarbanel Jan 30 '15 at 14:28

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, $(1 + a_1)\cdots (1 + a_n) > 1 + S$.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.