Mixing
A conjecture about power sum : $e^{ab}+e^{bc}+e^{ca}geq 3e^{sqrt{abc}}$ and $a+b+c=3$
$begingroup$
Hello I have this to propose :
Let $a,b,c$ be real positive numbers such that $a+b+c=3$ then we have :
$$e^{ab}+e^{bc}+e^{ca}geq 3e^{sqrt{abc}}$$
For a generalization I have this conjecture :
Let $a_i$ be $n$ real positive numbers such that $sum_{i=1}^{n}a_i=n$ then we have (with $a_{n+1}=a_1$):
$$sum_{i=1}^{n}e^{a_ia_{i+1}}geq ne^{Big(prod_{i=1}^{n}a_iBig)^{frac{1}{n-1}}}$$
In a first time I would like to know if there exists counter-examples and in a second time if it's true I would like some hints .
Thanks in advance .
real-analysis inequality exponential-function jensen-inequality
edited Feb 1 at 13:43
Martin R
30.8k33561
asked Feb 1 at 13:21
FatsWallersFatsWallers
1447
$endgroup$
add a comment |
$begingroup$
Hello I have this to propose :
Let $a,b,c$ be real positive numbers such that $a+b+c=3$ then we have :
$$e^{ab}+e^{bc}+e^{ca}geq 3e^{sqrt{abc}}$$
For a generalization I have this conjecture :
Let $a_i$ be $n$ real positive numbers such that $sum_{i=1}^{n}a_i=n$ then we have (with $a_{n+1}=a_1$):
$$sum_{i=1}^{n}e^{a_ia_{i+1}}geq ne^{Big(prod_{i=1}^{n}a_iBig)^{frac{1}{n-1}}}$$
In a first time I would like to know if there exists counter-examples and in a second time if it's true I would like some hints .
Thanks in advance .
real-analysis inequality exponential-function jensen-inequality
edited Feb 1 at 13:43
Martin R
30.8k33561
asked Feb 1 at 13:21
FatsWallersFatsWallers
1447
$endgroup$
$begingroup$
@MartinR Maybe OP is trying to create a contest-math problem.
$endgroup$
– yurnero
Feb 1 at 13:42
add a comment |
$begingroup$
Hello I have this to propose :
Let $a,b,c$ be real positive numbers such that $a+b+c=3$ then we have :
$$e^{ab}+e^{bc}+e^{ca}geq 3e^{sqrt{abc}}$$
For a generalization I have this conjecture :
Let $a_i$ be $n$ real positive numbers such that $sum_{i=1}^{n}a_i=n$ then we have (with $a_{n+1}=a_1$):
$$sum_{i=1}^{n}e^{a_ia_{i+1}}geq ne^{Big(prod_{i=1}^{n}a_iBig)^{frac{1}{n-1}}}$$
In a first time I would like to know if there exists counter-examples and in a second time if it's true I would like some hints .
Thanks in advance .
real-analysis inequality exponential-function jensen-inequality
edited Feb 1 at 13:43
Martin R
30.8k33561
asked Feb 1 at 13:21
FatsWallersFatsWallers
1447
$endgroup$
Hello I have this to propose :
Let $a,b,c$ be real positive numbers such that $a+b+c=3$ then we have :
$$e^{ab}+e^{bc}+e^{ca}geq 3e^{sqrt{abc}}$$
For a generalization I have this conjecture :
Let $a_i$ be $n$ real positive numbers such that $sum_{i=1}^{n}a_i=n$ then we have (with $a_{n+1}=a_1$):
$$sum_{i=1}^{n}e^{a_ia_{i+1}}geq ne^{Big(prod_{i=1}^{n}a_iBig)^{frac{1}{n-1}}}$$
In a first time I would like to know if there exists counter-examples and in a second time if it's true I would like some hints .
Thanks in advance .
real-analysis inequality exponential-function jensen-inequality
real-analysis inequality exponential-function jensen-inequality
edited Feb 1 at 13:43
Martin R
30.8k33561
asked Feb 1 at 13:21
FatsWallersFatsWallers
1447
edited Feb 1 at 13:43
Martin R
30.8k33561
asked Feb 1 at 13:21
FatsWallersFatsWallers
1447
edited Feb 1 at 13:43
Martin R
30.8k33561
edited Feb 1 at 13:43
Martin R
30.8k33561
edited Feb 1 at 13:43
Martin R
30.8k33561
30.8k33561
asked Feb 1 at 13:21
FatsWallersFatsWallers
1447
asked Feb 1 at 13:21
FatsWallersFatsWallers
1447
asked Feb 1 at 13:21
FatsWallersFatsWallers
1447
1447
$begingroup$
@MartinR Maybe OP is trying to create a contest-math problem.
$endgroup$
– yurnero
Feb 1 at 13:42
add a comment |
$begingroup$
@MartinR Maybe OP is trying to create a contest-math problem.
$endgroup$
– yurnero
Feb 1 at 13:42
$begingroup$
@MartinR Maybe OP is trying to create a contest-math problem.
$endgroup$
– yurnero
Feb 1 at 13:42
$begingroup$
@MartinR Maybe OP is trying to create a contest-math problem.
$endgroup$
– yurnero
Feb 1 at 13:42
add a comment |
2 Answers
2
active
oldest
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$begingroup$
By Jensen
$$e^{ab}+e^{ac}+e^{bc}geq3e^{frac{ab+ac+bc}{3}}geq3e^{sqrt{abc}}$$ because the last inequality it's
$$ab+ac+bcgeqsqrt{3(a+b+c)abc}$$ or after squaring of the both sides
$$sum_{cyc}c^2(a-b)^2geq0.$$
The second inequality is wrong.
Try $n=4$, $a_1=a_3=frac{1}{4}$ and $a_2=a_4=frac{7}{4}.$
answered Feb 1 at 13:38
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
add a comment |
$begingroup$
For the first, recall $xmapsto e^x$ is convex. Thus, by Jensen's inequality,
$$f(ab)+f(bc)+f(ca)geq 3f(frac{ab+bc+ca}{3})=3exp(frac{ab+bc+ca}{3}),$$
where $f(x)=e^x$.
Now, observe that, $(ab+bc+ca)^2geq 3abc(a+b+c)=9abc$. Hence, $frac{ab+bc+ca}{3}geq sqrt{abc}$. and therefore, using the fact that $f(cdot)$ is increasing, we conclude,
$$
e^{ab}+e^{bc}+e^{ca}geq 3f(frac{ab+bc+ca}{3})geq 3e^{sqrt{abc}}.
$$
For the second one, I am very confident that this approach should transfer without much hassle.
answered Feb 1 at 13:38
AaronAaron
2,015415
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
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active
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$begingroup$
By Jensen
$$e^{ab}+e^{ac}+e^{bc}geq3e^{frac{ab+ac+bc}{3}}geq3e^{sqrt{abc}}$$ because the last inequality it's
$$ab+ac+bcgeqsqrt{3(a+b+c)abc}$$ or after squaring of the both sides
$$sum_{cyc}c^2(a-b)^2geq0.$$
The second inequality is wrong.
Try $n=4$, $a_1=a_3=frac{1}{4}$ and $a_2=a_4=frac{7}{4}.$
answered Feb 1 at 13:38
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
add a comment |
$begingroup$
By Jensen
$$e^{ab}+e^{ac}+e^{bc}geq3e^{frac{ab+ac+bc}{3}}geq3e^{sqrt{abc}}$$ because the last inequality it's
$$ab+ac+bcgeqsqrt{3(a+b+c)abc}$$ or after squaring of the both sides
$$sum_{cyc}c^2(a-b)^2geq0.$$
The second inequality is wrong.
Try $n=4$, $a_1=a_3=frac{1}{4}$ and $a_2=a_4=frac{7}{4}.$
answered Feb 1 at 13:38
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
add a comment |
$begingroup$
By Jensen
$$e^{ab}+e^{ac}+e^{bc}geq3e^{frac{ab+ac+bc}{3}}geq3e^{sqrt{abc}}$$ because the last inequality it's
$$ab+ac+bcgeqsqrt{3(a+b+c)abc}$$ or after squaring of the both sides
$$sum_{cyc}c^2(a-b)^2geq0.$$
The second inequality is wrong.
Try $n=4$, $a_1=a_3=frac{1}{4}$ and $a_2=a_4=frac{7}{4}.$
answered Feb 1 at 13:38
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
By Jensen
$$e^{ab}+e^{ac}+e^{bc}geq3e^{frac{ab+ac+bc}{3}}geq3e^{sqrt{abc}}$$ because the last inequality it's
$$ab+ac+bcgeqsqrt{3(a+b+c)abc}$$ or after squaring of the both sides
$$sum_{cyc}c^2(a-b)^2geq0.$$
The second inequality is wrong.
Try $n=4$, $a_1=a_3=frac{1}{4}$ and $a_2=a_4=frac{7}{4}.$
answered Feb 1 at 13:38
Michael RozenbergMichael Rozenberg
110k1896201
edited Feb 1 at 14:43
answered Feb 1 at 13:38
Michael RozenbergMichael Rozenberg
110k1896201
answered Feb 1 at 13:38
Michael RozenbergMichael Rozenberg
110k1896201
answered Feb 1 at 13:38
Michael RozenbergMichael Rozenberg
110k1896201
110k1896201
add a comment |
add a comment |
$begingroup$
For the first, recall $xmapsto e^x$ is convex. Thus, by Jensen's inequality,
$$f(ab)+f(bc)+f(ca)geq 3f(frac{ab+bc+ca}{3})=3exp(frac{ab+bc+ca}{3}),$$
where $f(x)=e^x$.
Now, observe that, $(ab+bc+ca)^2geq 3abc(a+b+c)=9abc$. Hence, $frac{ab+bc+ca}{3}geq sqrt{abc}$. and therefore, using the fact that $f(cdot)$ is increasing, we conclude,
$$
e^{ab}+e^{bc}+e^{ca}geq 3f(frac{ab+bc+ca}{3})geq 3e^{sqrt{abc}}.
$$
For the second one, I am very confident that this approach should transfer without much hassle.
answered Feb 1 at 13:38
AaronAaron
2,015415
$endgroup$
add a comment |
$begingroup$
For the first, recall $xmapsto e^x$ is convex. Thus, by Jensen's inequality,
$$f(ab)+f(bc)+f(ca)geq 3f(frac{ab+bc+ca}{3})=3exp(frac{ab+bc+ca}{3}),$$
where $f(x)=e^x$.
Now, observe that, $(ab+bc+ca)^2geq 3abc(a+b+c)=9abc$. Hence, $frac{ab+bc+ca}{3}geq sqrt{abc}$. and therefore, using the fact that $f(cdot)$ is increasing, we conclude,
$$
e^{ab}+e^{bc}+e^{ca}geq 3f(frac{ab+bc+ca}{3})geq 3e^{sqrt{abc}}.
$$
For the second one, I am very confident that this approach should transfer without much hassle.
answered Feb 1 at 13:38
AaronAaron
2,015415
$endgroup$
add a comment |
$begingroup$
For the first, recall $xmapsto e^x$ is convex. Thus, by Jensen's inequality,
$$f(ab)+f(bc)+f(ca)geq 3f(frac{ab+bc+ca}{3})=3exp(frac{ab+bc+ca}{3}),$$
where $f(x)=e^x$.
Now, observe that, $(ab+bc+ca)^2geq 3abc(a+b+c)=9abc$. Hence, $frac{ab+bc+ca}{3}geq sqrt{abc}$. and therefore, using the fact that $f(cdot)$ is increasing, we conclude,
$$
e^{ab}+e^{bc}+e^{ca}geq 3f(frac{ab+bc+ca}{3})geq 3e^{sqrt{abc}}.
$$
For the second one, I am very confident that this approach should transfer without much hassle.
answered Feb 1 at 13:38
AaronAaron
2,015415
$endgroup$
For the first, recall $xmapsto e^x$ is convex. Thus, by Jensen's inequality,
$$f(ab)+f(bc)+f(ca)geq 3f(frac{ab+bc+ca}{3})=3exp(frac{ab+bc+ca}{3}),$$
where $f(x)=e^x$.
Now, observe that, $(ab+bc+ca)^2geq 3abc(a+b+c)=9abc$. Hence, $frac{ab+bc+ca}{3}geq sqrt{abc}$. and therefore, using the fact that $f(cdot)$ is increasing, we conclude,
$$
e^{ab}+e^{bc}+e^{ca}geq 3f(frac{ab+bc+ca}{3})geq 3e^{sqrt{abc}}.
$$
For the second one, I am very confident that this approach should transfer without much hassle.
answered Feb 1 at 13:38
AaronAaron
2,015415
answered Feb 1 at 13:38
AaronAaron
2,015415
answered Feb 1 at 13:38
AaronAaron
2,015415
answered Feb 1 at 13:38
AaronAaron
2,015415
2,015415
add a comment |
add a comment |
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$begingroup$
@MartinR Maybe OP is trying to create a contest-math problem.
$endgroup$
– yurnero
Feb 1 at 13:42