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I was trying to do this question. Let $a_1\times a_2\times a_3\times ....\times a_n = 1$ where all ai are positive real numbers. We have to prove that

$$(1+a_1)(1+a_2)....(1+a_n) \geq 2^n$$

How to go about proving this since we can't use $ AM\geq GM$? Although this inequality seems obvious.

Thanks

Sumanta
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nmnsharma007
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1 Answers1

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Since $x^2\geq0$ (square of a real number $\geq0$)

$(1-\sqrt{a_1})^2\geq 0 \implies 1+a_1\geq 2\sqrt{a_1}$

$(1-\sqrt{a_2})^2\geq 0 \implies 1+a_2\geq 2\sqrt{a_2}$

. . .

$(1-\sqrt{a_n})^2\geq 0 \implies 1+a_n\geq 2\sqrt{a_n}$

Then

$(1+a_1) \times (1+a_2) \times .... \times (1+a_n)= 2\sqrt{a_1} \times 2\sqrt{a_2} \times ...\times 2\sqrt{a_n} = 2^n \times \sqrt{a_1a_2...a_n}=2^n$

Lion Heart
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    First of all the inequalities are wrong $(1-\sqrt{a})^2\ge0\implies1+a\ge2\sqrt{a}$ not $2a$, as you have written. Also $1+a_\ge2\sqrt{a}$ is nothing but the AM-GM inequality! – QED Sep 27 '20 at 06:58
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    Oh thanks. I didn't notice. Is anything possible using induction then?? I can't find an inductive proof anywhere. – nmnsharma007 Sep 27 '20 at 07:04
  • @nmnsharma_007 I am going to try for induction that (per QED's comment) avoids even the hint of AM-GM. If I come up with something, I'll post a 2nd answer. In the meantime, I reversed the downvote. It could be argued that QED's criticism is still applicable, however. – user2661923 Sep 27 '20 at 07:10
  • https://cuhkmath.wordpress.com/2017/06/14/why-is-a%C2%B2-b%C2%B2-%E2%89%A5-2ab/ please check this one AM-GM is different – Lion Heart Sep 27 '20 at 07:13
  • What are you implying? – nmnsharma007 Sep 27 '20 at 07:24
  • Square of a number $ \geq 0$ – Lion Heart Sep 27 '20 at 07:27
  • Yeah,but, I feel like as QED said above, it can be argued that it uses AM-GM somehow. The answer Martin gave above, I need something like that. No trace of AM-GM at all.I might be wrong as well. – nmnsharma007 Sep 27 '20 at 07:35
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    @nmnsharma_007 I was going to try to craft an induction proof, but the question has been closed. However, I can offer this partial auxiliary to Lion Heart's answer that should satisfy QED's comment: Let $f(x) = 1 + x - 2(x)^{(1/2)} : x > 0.~$ Then $f'(x) = 1 - (x)^{(-1/2)}~$ and $~f''(x) = (1/2) \times (x)^{(-3/2)}.~$ Here we have that as $x \to 0^{+}, f(x) \to 1~$ and $~f(1) = 0.~$ The sole value for $x$ where $f'(x) = 0$ is $x=1~$ and $~f''(1) > 0.~$ Therefore $f(x)$ has an absolute minimum throughout $(0, \infty)~$ at $~x=1~$. Therefore for $~x > 0, ~(1 + x) \geq (x)^{(1/2)}.$ – user2661923 Sep 27 '20 at 07:57
  • @QED so there – user2661923 Sep 27 '20 at 07:59
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    That's calculus. You mentioned you would give inductive proof above. Why?Although Its good enough. Thanks. – nmnsharma007 Sep 27 '20 at 08:00
  • @nmnsharma_007 I used Calculus rather than any hint of AM-GM solely to satisfy QED's criticism. Further, ignoring how one concludes that $(1+x) \geq 2\sqrt{x}$, even if I had conjured an inductive proof for the remainder of the demonstration, I would still have regarded my conjuring as inferior to the remainder of Lion Heart's answer. Therefore, my sole motive in attempting induction on the remainder of the demonstration was simply to see if it was possible. – user2661923 Sep 27 '20 at 08:06
  • @nmnsharma_007 For the portion of the demonstration where I used Calculus, it was never my intention to attempt induction on that portion of the demonstration. – user2661923 Sep 27 '20 at 08:10