A nice power sum inequality
$begingroup$
Hello I have this theorem to propose :
Let $a_i$ be $n$ real positive numbers such that $sum_{i=1}^{n}a_i=n$ and $a_{n+1}=a_1$ then we have :
$$sum_{i=1}^{n}a_i^{a_i a_{i+1}+n}geq n$$
For $n=2$ you can apply Jensen's inequality and am-gm for two variables to get :
$$a^{ab+n}+b^{ab+n}geq 2(frac{a+b}{2})^{ab+n}geq 2(frac{a+b}{2})^{(frac{a+b}{2})^2+n}geq 2 $$ with $a+b=2$
So if we make the problem symmetric we get :
$$sum_{i=1}^{n}a_i^{a_i a_{i+1}+n}+sum_{i=1}^{n}a_{i+1}^{a_i a_{i+1}+n}geq 2n$$
But we have $n$ conditions instead of one condition .
So if you have a proof it would be nice .
Thanks in advance .
real-analysis inequality
asked Jan 18 at 13:13
FatsWallersFatsWallers
435
$endgroup$
add a comment |
$begingroup$
Hello I have this theorem to propose :
Let $a_i$ be $n$ real positive numbers such that $sum_{i=1}^{n}a_i=n$ and $a_{n+1}=a_1$ then we have :
$$sum_{i=1}^{n}a_i^{a_i a_{i+1}+n}geq n$$
For $n=2$ you can apply Jensen's inequality and am-gm for two variables to get :
$$a^{ab+n}+b^{ab+n}geq 2(frac{a+b}{2})^{ab+n}geq 2(frac{a+b}{2})^{(frac{a+b}{2})^2+n}geq 2 $$ with $a+b=2$
So if we make the problem symmetric we get :
$$sum_{i=1}^{n}a_i^{a_i a_{i+1}+n}+sum_{i=1}^{n}a_{i+1}^{a_i a_{i+1}+n}geq 2n$$
But we have $n$ conditions instead of one condition .
So if you have a proof it would be nice .
Thanks in advance .
real-analysis inequality
asked Jan 18 at 13:13
FatsWallersFatsWallers
435
$endgroup$
add a comment |
$begingroup$
Hello I have this theorem to propose :
Let $a_i$ be $n$ real positive numbers such that $sum_{i=1}^{n}a_i=n$ and $a_{n+1}=a_1$ then we have :
$$sum_{i=1}^{n}a_i^{a_i a_{i+1}+n}geq n$$
For $n=2$ you can apply Jensen's inequality and am-gm for two variables to get :
$$a^{ab+n}+b^{ab+n}geq 2(frac{a+b}{2})^{ab+n}geq 2(frac{a+b}{2})^{(frac{a+b}{2})^2+n}geq 2 $$ with $a+b=2$
So if we make the problem symmetric we get :
$$sum_{i=1}^{n}a_i^{a_i a_{i+1}+n}+sum_{i=1}^{n}a_{i+1}^{a_i a_{i+1}+n}geq 2n$$
But we have $n$ conditions instead of one condition .
So if you have a proof it would be nice .
Thanks in advance .
real-analysis inequality
asked Jan 18 at 13:13
FatsWallersFatsWallers
435
$endgroup$
Hello I have this theorem to propose :
Let $a_i$ be $n$ real positive numbers such that $sum_{i=1}^{n}a_i=n$ and $a_{n+1}=a_1$ then we have :
$$sum_{i=1}^{n}a_i^{a_i a_{i+1}+n}geq n$$
For $n=2$ you can apply Jensen's inequality and am-gm for two variables to get :
$$a^{ab+n}+b^{ab+n}geq 2(frac{a+b}{2})^{ab+n}geq 2(frac{a+b}{2})^{(frac{a+b}{2})^2+n}geq 2 $$ with $a+b=2$
So if we make the problem symmetric we get :
$$sum_{i=1}^{n}a_i^{a_i a_{i+1}+n}+sum_{i=1}^{n}a_{i+1}^{a_i a_{i+1}+n}geq 2n$$
But we have $n$ conditions instead of one condition .
So if you have a proof it would be nice .
Thanks in advance .
real-analysis inequality
real-analysis inequality
asked Jan 18 at 13:13
FatsWallersFatsWallers
435
asked Jan 18 at 13:13
FatsWallersFatsWallers
435
asked Jan 18 at 13:13
FatsWallersFatsWallers
435
asked Jan 18 at 13:13
FatsWallersFatsWallers
435
asked Jan 18 at 13:13
FatsWallersFatsWallers
435
435
add a comment |
add a comment |
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