Prove or disprove an inequality problem












0












$begingroup$


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











$endgroup$












  • $begingroup$
    Wouldn't the case where all $a_i$ are equal translate to proving $(1-(1-a)^n)(1+na)<1$?
    $endgroup$
    – Shubham Johri
    Nov 30 '18 at 17:22










  • $begingroup$
    Sorry, I edited the original question @ShubhamJohri
    $endgroup$
    – cscisgqr
    Nov 30 '18 at 17:26
















0












$begingroup$


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











$endgroup$












  • $begingroup$
    Wouldn't the case where all $a_i$ are equal translate to proving $(1-(1-a)^n)(1+na)<1$?
    $endgroup$
    – Shubham Johri
    Nov 30 '18 at 17:22










  • $begingroup$
    Sorry, I edited the original question @ShubhamJohri
    $endgroup$
    – cscisgqr
    Nov 30 '18 at 17:26














0












0








0





$begingroup$


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











$endgroup$




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 '18 at 18:05









user1551

72k566126




72k566126










asked Nov 30 '18 at 17:14









cscisgqrcscisgqr

42




42












  • $begingroup$
    Wouldn't the case where all $a_i$ are equal translate to proving $(1-(1-a)^n)(1+na)<1$?
    $endgroup$
    – Shubham Johri
    Nov 30 '18 at 17:22










  • $begingroup$
    Sorry, I edited the original question @ShubhamJohri
    $endgroup$
    – cscisgqr
    Nov 30 '18 at 17:26


















  • $begingroup$
    Wouldn't the case where all $a_i$ are equal translate to proving $(1-(1-a)^n)(1+na)<1$?
    $endgroup$
    – Shubham Johri
    Nov 30 '18 at 17:22










  • $begingroup$
    Sorry, I edited the original question @ShubhamJohri
    $endgroup$
    – cscisgqr
    Nov 30 '18 at 17:26
















$begingroup$
Wouldn't the case where all $a_i$ are equal translate to proving $(1-(1-a)^n)(1+na)<1$?
$endgroup$
– Shubham Johri
Nov 30 '18 at 17:22




$begingroup$
Wouldn't the case where all $a_i$ are equal translate to proving $(1-(1-a)^n)(1+na)<1$?
$endgroup$
– Shubham Johri
Nov 30 '18 at 17:22












$begingroup$
Sorry, I edited the original question @ShubhamJohri
$endgroup$
– cscisgqr
Nov 30 '18 at 17:26




$begingroup$
Sorry, I edited the original question @ShubhamJohri
$endgroup$
– cscisgqr
Nov 30 '18 at 17:26










2 Answers
2






active

oldest

votes


















2












$begingroup$

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











$endgroup$













  • $begingroup$
    Sorry, I edited the original question
    $endgroup$
    – cscisgqr
    Nov 30 '18 at 17:22










  • $begingroup$
    @cscisgqr I also fixed. See now.
    $endgroup$
    – Michael Rozenberg
    Nov 30 '18 at 17:34










  • $begingroup$
    How did numerator become n+1? @Michael Rozenberg
    $endgroup$
    – cscisgqr
    Nov 30 '18 at 17:47










  • $begingroup$
    @cscisgqr It's $1+sumlimits_{i=1}^n1=1+n$ and $-sumlimits_{i=1}^na_i+sumlimits_{i=1}^na_i=0.$
    $endgroup$
    – Michael Rozenberg
    Nov 30 '18 at 17:50








  • 1




    $begingroup$
    Nice application of AM-GM +1
    $endgroup$
    – Macavity
    Nov 30 '18 at 17:56



















1












$begingroup$

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









$endgroup$













  • $begingroup$
    Beautiful solution! +1.
    $endgroup$
    – Michael Rozenberg
    Nov 30 '18 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












$begingroup$

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











$endgroup$













  • $begingroup$
    Sorry, I edited the original question
    $endgroup$
    – cscisgqr
    Nov 30 '18 at 17:22










  • $begingroup$
    @cscisgqr I also fixed. See now.
    $endgroup$
    – Michael Rozenberg
    Nov 30 '18 at 17:34










  • $begingroup$
    How did numerator become n+1? @Michael Rozenberg
    $endgroup$
    – cscisgqr
    Nov 30 '18 at 17:47










  • $begingroup$
    @cscisgqr It's $1+sumlimits_{i=1}^n1=1+n$ and $-sumlimits_{i=1}^na_i+sumlimits_{i=1}^na_i=0.$
    $endgroup$
    – Michael Rozenberg
    Nov 30 '18 at 17:50








  • 1




    $begingroup$
    Nice application of AM-GM +1
    $endgroup$
    – Macavity
    Nov 30 '18 at 17:56
















2












$begingroup$

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











$endgroup$













  • $begingroup$
    Sorry, I edited the original question
    $endgroup$
    – cscisgqr
    Nov 30 '18 at 17:22










  • $begingroup$
    @cscisgqr I also fixed. See now.
    $endgroup$
    – Michael Rozenberg
    Nov 30 '18 at 17:34










  • $begingroup$
    How did numerator become n+1? @Michael Rozenberg
    $endgroup$
    – cscisgqr
    Nov 30 '18 at 17:47










  • $begingroup$
    @cscisgqr It's $1+sumlimits_{i=1}^n1=1+n$ and $-sumlimits_{i=1}^na_i+sumlimits_{i=1}^na_i=0.$
    $endgroup$
    – Michael Rozenberg
    Nov 30 '18 at 17:50








  • 1




    $begingroup$
    Nice application of AM-GM +1
    $endgroup$
    – Macavity
    Nov 30 '18 at 17:56














2












2








2





$begingroup$

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











$endgroup$



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 '18 at 17:57

























answered Nov 30 '18 at 17:22









Michael RozenbergMichael Rozenberg

97.8k1590188




97.8k1590188












  • $begingroup$
    Sorry, I edited the original question
    $endgroup$
    – cscisgqr
    Nov 30 '18 at 17:22










  • $begingroup$
    @cscisgqr I also fixed. See now.
    $endgroup$
    – Michael Rozenberg
    Nov 30 '18 at 17:34










  • $begingroup$
    How did numerator become n+1? @Michael Rozenberg
    $endgroup$
    – cscisgqr
    Nov 30 '18 at 17:47










  • $begingroup$
    @cscisgqr It's $1+sumlimits_{i=1}^n1=1+n$ and $-sumlimits_{i=1}^na_i+sumlimits_{i=1}^na_i=0.$
    $endgroup$
    – Michael Rozenberg
    Nov 30 '18 at 17:50








  • 1




    $begingroup$
    Nice application of AM-GM +1
    $endgroup$
    – Macavity
    Nov 30 '18 at 17:56


















  • $begingroup$
    Sorry, I edited the original question
    $endgroup$
    – cscisgqr
    Nov 30 '18 at 17:22










  • $begingroup$
    @cscisgqr I also fixed. See now.
    $endgroup$
    – Michael Rozenberg
    Nov 30 '18 at 17:34










  • $begingroup$
    How did numerator become n+1? @Michael Rozenberg
    $endgroup$
    – cscisgqr
    Nov 30 '18 at 17:47










  • $begingroup$
    @cscisgqr It's $1+sumlimits_{i=1}^n1=1+n$ and $-sumlimits_{i=1}^na_i+sumlimits_{i=1}^na_i=0.$
    $endgroup$
    – Michael Rozenberg
    Nov 30 '18 at 17:50








  • 1




    $begingroup$
    Nice application of AM-GM +1
    $endgroup$
    – Macavity
    Nov 30 '18 at 17:56
















$begingroup$
Sorry, I edited the original question
$endgroup$
– cscisgqr
Nov 30 '18 at 17:22




$begingroup$
Sorry, I edited the original question
$endgroup$
– cscisgqr
Nov 30 '18 at 17:22












$begingroup$
@cscisgqr I also fixed. See now.
$endgroup$
– Michael Rozenberg
Nov 30 '18 at 17:34




$begingroup$
@cscisgqr I also fixed. See now.
$endgroup$
– Michael Rozenberg
Nov 30 '18 at 17:34












$begingroup$
How did numerator become n+1? @Michael Rozenberg
$endgroup$
– cscisgqr
Nov 30 '18 at 17:47




$begingroup$
How did numerator become n+1? @Michael Rozenberg
$endgroup$
– cscisgqr
Nov 30 '18 at 17:47












$begingroup$
@cscisgqr It's $1+sumlimits_{i=1}^n1=1+n$ and $-sumlimits_{i=1}^na_i+sumlimits_{i=1}^na_i=0.$
$endgroup$
– Michael Rozenberg
Nov 30 '18 at 17:50






$begingroup$
@cscisgqr It's $1+sumlimits_{i=1}^n1=1+n$ and $-sumlimits_{i=1}^na_i+sumlimits_{i=1}^na_i=0.$
$endgroup$
– Michael Rozenberg
Nov 30 '18 at 17:50






1




1




$begingroup$
Nice application of AM-GM +1
$endgroup$
– Macavity
Nov 30 '18 at 17:56




$begingroup$
Nice application of AM-GM +1
$endgroup$
– Macavity
Nov 30 '18 at 17:56











1












$begingroup$

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









$endgroup$













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
















1












$begingroup$

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









$endgroup$













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














1












1








1





$begingroup$

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









$endgroup$



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 '18 at 18:00









Shubham JohriShubham Johri

4,666717




4,666717












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


















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
















$begingroup$
Beautiful solution! +1.
$endgroup$
– Michael Rozenberg
Nov 30 '18 at 18:02




$begingroup$
Beautiful solution! +1.
$endgroup$
– Michael Rozenberg
Nov 30 '18 at 18:02


















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