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How could you prove these inequalities wihout induction:($a_k$ are non-negative)

1)$\prod_{k=1}^n(1+a_k)\ge1+\sum_{k=1}^n a_k$

2)$\prod_{k=1}^n(1+a_k)\le1+\frac{\sum_{k=1}^na_k}{1!}+\ldots+\frac{(\sum_{k=1}^na_k)^n}{n!}$

3)$\prod_{k=1}^n(1+a_k)\le\frac1{1-\sum_{k=1}^na_k}, \ \forall\sum_{k=1}^na_k\lt1$

I did not get any positive result by the use of AM-GM inequalities. Induction proves it, but is a little longer for 2). As for 3) the geometric series seems the way. Any hints. Thanks beforehand.

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    The first is obvious. – 2017-01-19
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    @MichaelRozenberg how come it is obvious? – 2017-01-19
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    @MichaelRozenberg ok, multiplying termwise gives a positive increment in LHS, isnt it? – 2017-01-19

1 Answers 1

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1) $$\prod_{k=1}^n(1+a_k)=1+\sum_{k=1}^nx_k+\sum_{1\leq iMaclaurin's inequality 3) follows from 2) immediately:

Let $a_1+a_2+...+a_n=x$...

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    by the way how does the maclaurin's inequality help? it is talking about symmetric terms of products whereas I only have sum on RHS, isnt it? – 2017-01-19
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    Because, for example $\frac{\sum\limits_{1\leq i – 2017-01-19
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    thanks, but by the way the inequalities of maclaurin, newton's and muirhead's etc. are not in any standard course of analysis or algebra, how did you manage to learn it? – 2017-01-19
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    @vidyarthi In 1979 I found very nice and easy proof of the Maclaurin. I am ready to show. Do you want to see my proof? – 2017-01-19
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    yes, I am interested. – 2017-01-20
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    @vidyarthi See here http://math.stackexchange.com/questions/197955/how-to-prove-this-inequality-sqrt-fracabbccddaacbd6-geq-sqrt3/1596047#1596047 This idea we can use in the general case. – 2017-01-20