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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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    $\begingroup$ The first is obvious. $\endgroup$ Commented Jan 19, 2017 at 9:03
  • $\begingroup$ @MichaelRozenberg how come it is obvious? $\endgroup$
    – vidyarthi
    Commented Jan 19, 2017 at 9:05
  • $\begingroup$ @MichaelRozenberg ok, multiplying termwise gives a positive increment in LHS, isnt it? $\endgroup$
    – vidyarthi
    Commented Jan 19, 2017 at 9:07

1 Answer 1

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1) $$\prod_{k=1}^n(1+a_k)=1+\sum_{k=1}^nx_k+\sum_{1\leq i<j\leq n}a_ia_j+...\geq1+\sum_{k=1}^na_k$$ 2) follows from Maclaurin's inequality 3) follows from 2) immediately:

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

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  • $\begingroup$ 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? $\endgroup$
    – vidyarthi
    Commented Jan 19, 2017 at 9:15
  • $\begingroup$ Because, for example $\frac{\sum\limits_{1\leq i<j\leq n}a_ia_j}{\binom{n}{2}}\leq\left(\frac{a_1+...+a_n}{n}\right)^2$ $\endgroup$ Commented Jan 19, 2017 at 9:19
  • $\begingroup$ 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? $\endgroup$
    – vidyarthi
    Commented Jan 19, 2017 at 9:29
  • $\begingroup$ @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? $\endgroup$ Commented Jan 19, 2017 at 12:31
  • $\begingroup$ yes, I am interested. $\endgroup$
    – vidyarthi
    Commented Jan 20, 2017 at 6:07

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