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Inequality with log-sum

Impossible inequality : Dottie NumberA simple application of Jensen's inequalityTitan inequality for amateurInequality for amateursKaramata + Jensen = Niculescu's inequality (version of 1991)Hard inequality with condition ($xyz=1$)Inequality using JensenRefinement of a strong inequalityPower sum inequalityPower sum refinement inequality

1

$begingroup$

I'm interested by the following problem :

Let $a,b,c,d>0$ such that $a+b+c+d=4$ then we have :
$$abln(a)+bcln(b)+cdln(c)+daln(d)> -1.37$$

My try for $bgeq 3,aleq1,cleq 1,dleq1$:

We have the following inequality :

Let $varepsilon >0$ and $x>0$ then we have :
$$fracx(1-varepsilon)varepsilon-frac(x)^1-varepsilonvarepsilon e^varepsilonleq xln(x)$$

So we have to prove (here I take $varepsilon=0.001$):

$$fracab(1-varepsilon)varepsilon-fracb(a)^1-varepsilonvarepsilon e^varepsilon+fracbc(1-varepsilon)varepsilon-fracc(b)^1-varepsilonvarepsilon e^varepsilon+fraccd(1-varepsilon)varepsilon-fracd(c)^1-varepsilonvarepsilon e^varepsilon+fracda(1-varepsilon)varepsilon-fraca(d)^1-varepsilonvarepsilon e^varepsilon>-1.37$$

Now I use a form of Jensen's inequality :

Let $a,b,c,d>0$ such that $a+b+c+d=4$ and $bgeq 3,aleq1,cleq 1,dleq1$ then we have :
$$3.7(fracab+bc+cd+da3.7)^1-varepsilon>b(a)^1-varepsilon+c(b)^1-varepsilon+d(c)^1-varepsilon+a(d)^1-varepsilon$$

If we combine this two fact we have to prove :
$$frac(ab+bc+cd+da)(1-varepsilon)varepsilon-frac3.7(fracab+bc+cd+da3.7)^1-varepsilonvarepsilon e^varepsilon>-1.37$$

Wich is a one variable inequality because $$varepsilon=0.001$$

I have two questions :

How to complete my reasoning ?
How to prove this version of Jensen's inequality ?

Thanks in advance for your time .

real-analysis logarithms jensen-inequality

share|cite|improve this question

asked Mar 21 at 7:57

max8128max8128

231422

$endgroup$

add a comment | 

1

$begingroup$

I'm interested by the following problem :

Let $a,b,c,d>0$ such that $a+b+c+d=4$ then we have :
$$abln(a)+bcln(b)+cdln(c)+daln(d)> -1.37$$

My try for $bgeq 3,aleq1,cleq 1,dleq1$:

We have the following inequality :

Let $varepsilon >0$ and $x>0$ then we have :
$$fracx(1-varepsilon)varepsilon-frac(x)^1-varepsilonvarepsilon e^varepsilonleq xln(x)$$

So we have to prove (here I take $varepsilon=0.001$):

$$fracab(1-varepsilon)varepsilon-fracb(a)^1-varepsilonvarepsilon e^varepsilon+fracbc(1-varepsilon)varepsilon-fracc(b)^1-varepsilonvarepsilon e^varepsilon+fraccd(1-varepsilon)varepsilon-fracd(c)^1-varepsilonvarepsilon e^varepsilon+fracda(1-varepsilon)varepsilon-fraca(d)^1-varepsilonvarepsilon e^varepsilon>-1.37$$

Now I use a form of Jensen's inequality :

Let $a,b,c,d>0$ such that $a+b+c+d=4$ and $bgeq 3,aleq1,cleq 1,dleq1$ then we have :
$$3.7(fracab+bc+cd+da3.7)^1-varepsilon>b(a)^1-varepsilon+c(b)^1-varepsilon+d(c)^1-varepsilon+a(d)^1-varepsilon$$

If we combine this two fact we have to prove :
$$frac(ab+bc+cd+da)(1-varepsilon)varepsilon-frac3.7(fracab+bc+cd+da3.7)^1-varepsilonvarepsilon e^varepsilon>-1.37$$

Wich is a one variable inequality because $$varepsilon=0.001$$

I have two questions :

How to complete my reasoning ?
How to prove this version of Jensen's inequality ?

Thanks in advance for your time .

real-analysis logarithms jensen-inequality

share|cite|improve this question

asked Mar 21 at 7:57

max8128max8128

231422

$endgroup$

add a comment | 

$begingroup$

I'm interested by the following problem :

Let $a,b,c,d>0$ such that $a+b+c+d=4$ then we have :
$$abln(a)+bcln(b)+cdln(c)+daln(d)> -1.37$$

My try for $bgeq 3,aleq1,cleq 1,dleq1$:

We have the following inequality :

Let $varepsilon >0$ and $x>0$ then we have :
$$fracx(1-varepsilon)varepsilon-frac(x)^1-varepsilonvarepsilon e^varepsilonleq xln(x)$$

So we have to prove (here I take $varepsilon=0.001$):

$$fracab(1-varepsilon)varepsilon-fracb(a)^1-varepsilonvarepsilon e^varepsilon+fracbc(1-varepsilon)varepsilon-fracc(b)^1-varepsilonvarepsilon e^varepsilon+fraccd(1-varepsilon)varepsilon-fracd(c)^1-varepsilonvarepsilon e^varepsilon+fracda(1-varepsilon)varepsilon-fraca(d)^1-varepsilonvarepsilon e^varepsilon>-1.37$$

Now I use a form of Jensen's inequality :

Let $a,b,c,d>0$ such that $a+b+c+d=4$ and $bgeq 3,aleq1,cleq 1,dleq1$ then we have :
$$3.7(fracab+bc+cd+da3.7)^1-varepsilon>b(a)^1-varepsilon+c(b)^1-varepsilon+d(c)^1-varepsilon+a(d)^1-varepsilon$$

If we combine this two fact we have to prove :
$$frac(ab+bc+cd+da)(1-varepsilon)varepsilon-frac3.7(fracab+bc+cd+da3.7)^1-varepsilonvarepsilon e^varepsilon>-1.37$$

Wich is a one variable inequality because $$varepsilon=0.001$$

I have two questions :

How to complete my reasoning ?
How to prove this version of Jensen's inequality ?

Thanks in advance for your time .

real-analysis logarithms jensen-inequality

share|cite|improve this question

asked Mar 21 at 7:57

max8128max8128

231422

$endgroup$

share|cite|improve this question

asked Mar 21 at 7:57

max8128max8128

231422

share|cite|improve this question

asked Mar 21 at 7:57

max8128max8128

231422

asked Mar 21 at 7:57

max8128max8128

231422

asked Mar 21 at 7:57

max8128max8128

231422