Prove or disprove an inequality problem












0














Let $ngeq 1$ be an integer and let $a_1,ldots,a_n$ be positive real numbers, all between $0$ and $1$.



Is it possible to prove or disprove:
$$
{(prod_{i=1}^{n}(1-a_i))}{(1+sum_{i=1}^{n}a_i)}<1
$$
To prove this, I was able to prove this to be true if all $a_1,ldots,a_n$ are the same which is$(1-a)^n(1+an)<1
$ using Bernoulli's Inequality:



Take $x$ root of the inequality:
$$(1-a)(1+an)^{1/a}leqslant (1-n)(1+n)=1-n^2<1$$
But what if $a_1,ldots,a_n$ do not have uniform values?






inequality a.m.-g.m.-inequality







share|cite|improve this question
























  • Wouldn't the case where all $a_i$ are equal translate to proving $(1-(1-a)^n)(1+na)<1$?
    – Shubham Johri
    Nov 30 at 17:22










  • Sorry, I edited the original question @ShubhamJohri
    – cscisgqr
    Nov 30 at 17:26
















0














Let $ngeq 1$ be an integer and let $a_1,ldots,a_n$ be positive real numbers, all between $0$ and $1$.



Is it possible to prove or disprove:
$$
{(prod_{i=1}^{n}(1-a_i))}{(1+sum_{i=1}^{n}a_i)}<1
$$
To prove this, I was able to prove this to be true if all $a_1,ldots,a_n$ are the same which is$(1-a)^n(1+an)<1
$ using Bernoulli's Inequality:



Take $x$ root of the inequality:
$$(1-a)(1+an)^{1/a}leqslant (1-n)(1+n)=1-n^2<1$$
But what if $a_1,ldots,a_n$ do not have uniform values?






inequality a.m.-g.m.-inequality







share|cite|improve this question
























  • Wouldn't the case where all $a_i$ are equal translate to proving $(1-(1-a)^n)(1+na)<1$?
    – Shubham Johri
    Nov 30 at 17:22










  • Sorry, I edited the original question @ShubhamJohri
    – cscisgqr
    Nov 30 at 17:26














0












0








0







Let $ngeq 1$ be an integer and let $a_1,ldots,a_n$ be positive real numbers, all between $0$ and $1$.



Is it possible to prove or disprove:
$$
{(prod_{i=1}^{n}(1-a_i))}{(1+sum_{i=1}^{n}a_i)}<1
$$
To prove this, I was able to prove this to be true if all $a_1,ldots,a_n$ are the same which is$(1-a)^n(1+an)<1
$ using Bernoulli's Inequality:



Take $x$ root of the inequality:
$$(1-a)(1+an)^{1/a}leqslant (1-n)(1+n)=1-n^2<1$$
But what if $a_1,ldots,a_n$ do not have uniform values?






inequality a.m.-g.m.-inequality







share|cite|improve this question















Let $ngeq 1$ be an integer and let $a_1,ldots,a_n$ be positive real numbers, all between $0$ and $1$.



Is it possible to prove or disprove:
$$
{(prod_{i=1}^{n}(1-a_i))}{(1+sum_{i=1}^{n}a_i)}<1
$$
To prove this, I was able to prove this to be true if all $a_1,ldots,a_n$ are the same which is$(1-a)^n(1+an)<1
$ using Bernoulli's Inequality:



Take $x$ root of the inequality:
$$(1-a)(1+an)^{1/a}leqslant (1-n)(1+n)=1-n^2<1$$
But what if $a_1,ldots,a_n$ do not have uniform values?






inequality a.m.-g.m.-inequality




inequality a.m.-g.m.-inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 at 18:05









user1551

71.2k566125




71.2k566125










asked Nov 30 at 17:14









cscisgqr

42




42












  • Wouldn't the case where all $a_i$ are equal translate to proving $(1-(1-a)^n)(1+na)<1$?
    – Shubham Johri
    Nov 30 at 17:22










  • Sorry, I edited the original question @ShubhamJohri
    – cscisgqr
    Nov 30 at 17:26


















  • Wouldn't the case where all $a_i$ are equal translate to proving $(1-(1-a)^n)(1+na)<1$?
    – Shubham Johri
    Nov 30 at 17:22










  • Sorry, I edited the original question @ShubhamJohri
    – cscisgqr
    Nov 30 at 17:26
















Wouldn't the case where all $a_i$ are equal translate to proving $(1-(1-a)^n)(1+na)<1$?
– Shubham Johri
Nov 30 at 17:22




Wouldn't the case where all $a_i$ are equal translate to proving $(1-(1-a)^n)(1+na)<1$?
– Shubham Johri
Nov 30 at 17:22












Sorry, I edited the original question @ShubhamJohri
– cscisgqr
Nov 30 at 17:26




Sorry, I edited the original question @ShubhamJohri
– cscisgqr
Nov 30 at 17:26










2 Answers
2






active

oldest

votes


















2














By AM-GM
$$prod_{i=1}^n(1-a_i)left(1+sum_{i=1}^na_iright)<left(frac{sumlimits_{i=1}^n(1-a_i)+1+sumlimits_{i=1}^na_i}{n+1}right)^{n+1}=left(frac{n+1}{n+1}right)^{n+1}=1.$$
The equality does not occur because $1-a_i<1+sumlimits_{i=1}^na_i.$






share|cite|improve this answer























  • Sorry, I edited the original question
    – cscisgqr
    Nov 30 at 17:22










  • @cscisgqr I also fixed. See now.
    – Michael Rozenberg
    Nov 30 at 17:34










  • How did numerator become n+1? @Michael Rozenberg
    – cscisgqr
    Nov 30 at 17:47










  • @cscisgqr It's $1+sumlimits_{i=1}^n1=1+n$ and $-sumlimits_{i=1}^na_i+sumlimits_{i=1}^na_i=0.$
    – Michael Rozenberg
    Nov 30 at 17:50








  • 1




    Nice application of AM-GM +1
    – Macavity
    Nov 30 at 17:56



















1














Note that $1+sum_1^na_i<prod_1^n(1+a_i)$



$implies (prod_1^n(1-a_i))(1+sum_1^na_i)<(prod_1^n(1-a_i))(prod_1^n(1+a_i))=prod_1^n(1-a_i^2)<1$






share|cite|improve this answer





















  • Beautiful solution! +1.
    – Michael Rozenberg
    Nov 30 at 18:02











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














By AM-GM
$$prod_{i=1}^n(1-a_i)left(1+sum_{i=1}^na_iright)<left(frac{sumlimits_{i=1}^n(1-a_i)+1+sumlimits_{i=1}^na_i}{n+1}right)^{n+1}=left(frac{n+1}{n+1}right)^{n+1}=1.$$
The equality does not occur because $1-a_i<1+sumlimits_{i=1}^na_i.$






share|cite|improve this answer























  • Sorry, I edited the original question
    – cscisgqr
    Nov 30 at 17:22










  • @cscisgqr I also fixed. See now.
    – Michael Rozenberg
    Nov 30 at 17:34










  • How did numerator become n+1? @Michael Rozenberg
    – cscisgqr
    Nov 30 at 17:47










  • @cscisgqr It's $1+sumlimits_{i=1}^n1=1+n$ and $-sumlimits_{i=1}^na_i+sumlimits_{i=1}^na_i=0.$
    – Michael Rozenberg
    Nov 30 at 17:50








  • 1




    Nice application of AM-GM +1
    – Macavity
    Nov 30 at 17:56
















2














By AM-GM
$$prod_{i=1}^n(1-a_i)left(1+sum_{i=1}^na_iright)<left(frac{sumlimits_{i=1}^n(1-a_i)+1+sumlimits_{i=1}^na_i}{n+1}right)^{n+1}=left(frac{n+1}{n+1}right)^{n+1}=1.$$
The equality does not occur because $1-a_i<1+sumlimits_{i=1}^na_i.$






share|cite|improve this answer























  • Sorry, I edited the original question
    – cscisgqr
    Nov 30 at 17:22










  • @cscisgqr I also fixed. See now.
    – Michael Rozenberg
    Nov 30 at 17:34










  • How did numerator become n+1? @Michael Rozenberg
    – cscisgqr
    Nov 30 at 17:47










  • @cscisgqr It's $1+sumlimits_{i=1}^n1=1+n$ and $-sumlimits_{i=1}^na_i+sumlimits_{i=1}^na_i=0.$
    – Michael Rozenberg
    Nov 30 at 17:50








  • 1




    Nice application of AM-GM +1
    – Macavity
    Nov 30 at 17:56














2












2








2






By AM-GM
$$prod_{i=1}^n(1-a_i)left(1+sum_{i=1}^na_iright)<left(frac{sumlimits_{i=1}^n(1-a_i)+1+sumlimits_{i=1}^na_i}{n+1}right)^{n+1}=left(frac{n+1}{n+1}right)^{n+1}=1.$$
The equality does not occur because $1-a_i<1+sumlimits_{i=1}^na_i.$






share|cite|improve this answer














By AM-GM
$$prod_{i=1}^n(1-a_i)left(1+sum_{i=1}^na_iright)<left(frac{sumlimits_{i=1}^n(1-a_i)+1+sumlimits_{i=1}^na_i}{n+1}right)^{n+1}=left(frac{n+1}{n+1}right)^{n+1}=1.$$
The equality does not occur because $1-a_i<1+sumlimits_{i=1}^na_i.$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 30 at 17:57

























answered Nov 30 at 17:22









Michael Rozenberg

95.4k1588183




95.4k1588183












  • Sorry, I edited the original question
    – cscisgqr
    Nov 30 at 17:22










  • @cscisgqr I also fixed. See now.
    – Michael Rozenberg
    Nov 30 at 17:34










  • How did numerator become n+1? @Michael Rozenberg
    – cscisgqr
    Nov 30 at 17:47










  • @cscisgqr It's $1+sumlimits_{i=1}^n1=1+n$ and $-sumlimits_{i=1}^na_i+sumlimits_{i=1}^na_i=0.$
    – Michael Rozenberg
    Nov 30 at 17:50








  • 1




    Nice application of AM-GM +1
    – Macavity
    Nov 30 at 17:56


















  • Sorry, I edited the original question
    – cscisgqr
    Nov 30 at 17:22










  • @cscisgqr I also fixed. See now.
    – Michael Rozenberg
    Nov 30 at 17:34










  • How did numerator become n+1? @Michael Rozenberg
    – cscisgqr
    Nov 30 at 17:47










  • @cscisgqr It's $1+sumlimits_{i=1}^n1=1+n$ and $-sumlimits_{i=1}^na_i+sumlimits_{i=1}^na_i=0.$
    – Michael Rozenberg
    Nov 30 at 17:50








  • 1




    Nice application of AM-GM +1
    – Macavity
    Nov 30 at 17:56
















Sorry, I edited the original question
– cscisgqr
Nov 30 at 17:22




Sorry, I edited the original question
– cscisgqr
Nov 30 at 17:22












@cscisgqr I also fixed. See now.
– Michael Rozenberg
Nov 30 at 17:34




@cscisgqr I also fixed. See now.
– Michael Rozenberg
Nov 30 at 17:34












How did numerator become n+1? @Michael Rozenberg
– cscisgqr
Nov 30 at 17:47




How did numerator become n+1? @Michael Rozenberg
– cscisgqr
Nov 30 at 17:47












@cscisgqr It's $1+sumlimits_{i=1}^n1=1+n$ and $-sumlimits_{i=1}^na_i+sumlimits_{i=1}^na_i=0.$
– Michael Rozenberg
Nov 30 at 17:50






@cscisgqr It's $1+sumlimits_{i=1}^n1=1+n$ and $-sumlimits_{i=1}^na_i+sumlimits_{i=1}^na_i=0.$
– Michael Rozenberg
Nov 30 at 17:50






1




1




Nice application of AM-GM +1
– Macavity
Nov 30 at 17:56




Nice application of AM-GM +1
– Macavity
Nov 30 at 17:56











1














Note that $1+sum_1^na_i<prod_1^n(1+a_i)$



$implies (prod_1^n(1-a_i))(1+sum_1^na_i)<(prod_1^n(1-a_i))(prod_1^n(1+a_i))=prod_1^n(1-a_i^2)<1$






share|cite|improve this answer





















  • Beautiful solution! +1.
    – Michael Rozenberg
    Nov 30 at 18:02
















1














Note that $1+sum_1^na_i<prod_1^n(1+a_i)$



$implies (prod_1^n(1-a_i))(1+sum_1^na_i)<(prod_1^n(1-a_i))(prod_1^n(1+a_i))=prod_1^n(1-a_i^2)<1$






share|cite|improve this answer





















  • Beautiful solution! +1.
    – Michael Rozenberg
    Nov 30 at 18:02














1












1








1






Note that $1+sum_1^na_i<prod_1^n(1+a_i)$



$implies (prod_1^n(1-a_i))(1+sum_1^na_i)<(prod_1^n(1-a_i))(prod_1^n(1+a_i))=prod_1^n(1-a_i^2)<1$






share|cite|improve this answer












Note that $1+sum_1^na_i<prod_1^n(1+a_i)$



$implies (prod_1^n(1-a_i))(1+sum_1^na_i)<(prod_1^n(1-a_i))(prod_1^n(1+a_i))=prod_1^n(1-a_i^2)<1$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 30 at 18:00









Shubham Johri

3,791716




3,791716












  • Beautiful solution! +1.
    – Michael Rozenberg
    Nov 30 at 18:02


















  • Beautiful solution! +1.
    – Michael Rozenberg
    Nov 30 at 18:02
















Beautiful solution! +1.
– Michael Rozenberg
Nov 30 at 18:02




Beautiful solution! +1.
– Michael Rozenberg
Nov 30 at 18:02


















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