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Help to prove that an inequality holds

  • #1

Homework Statement



I am trying to prove that

[tex]\frac{1}{512} \left[101-(1-\alpha ) \gamma ^n \left(2 (37+64 \alpha )+27 (1-\alpha ) \gamma ^n\right) (1-\delta )-\frac{128 n (1-\alpha ) \alpha (1-\gamma ) \gamma ^{-1+n} (1-\delta )}{\alpha +\delta -\alpha \delta }\right]\geq0[/tex]

for

[tex]\alpha,\delta,\gamma\in(0,1)[/tex]

and

[tex]n\geq2[/tex].

Edit: I should add that this isn't homework per se, but rather part of my own ongoing research project. As such, it may be that the inequality does not hold over this range of parameters. (If someone can find a counterexample, that would also help). Playing with this inequality in Mathematica's manipulate function, however, it seems that it does indeed hold over the relevant range.

Any help with this is warmly appreciated.

Homework Equations



So far, I have managed to reduce this problem to one of showing that

[tex]\gamma ^{n-1} \left(128 n (1-\gamma )+\gamma \left(74+27 \gamma ^n\right)\right)\leq101[/tex]

for

[tex]\gamma\in(0,1)[/tex]

and

[tex]n\geq2[/tex]

(for details see below).

The Attempt at a Solution



The second term in the big square brackets is decreasing in (absolute) size in [tex]\delta[/tex]. Taking the derrivative of the third term inside the square brackets yields

[tex]\frac{\partial}{\partial\delta}\frac{128 n (1-\alpha ) \alpha (1-\gamma ) \gamma ^{-1+n} (1-\delta )}{\alpha +\delta -\alpha \delta }=-\frac{128 n (1-\alpha ) \alpha (1-\gamma ) \gamma ^{-1+n}}{(\alpha +\delta -\alpha \delta )^2}\leq0[/tex].

The whole LHS of the inequality is therefore increasing in [tex]\delta[/tex], so if the inequality holds for [tex]\delta=0[/tex], it holds for all [tex]\delta\in(0,1)[/tex]. Let's take the limit as [tex]\delta[/tex] approaches zero. The problem is now reduced to one of showing that

[tex]101 \geq (1-\alpha)\gamma^{n-1}\left(128n(1-\gamma)+\gamma\left(2(37+64\alpha)+27(1-\alpha)\gamma^n\right)\right)[/tex]

Differentiating the R.H.S. of this W.R.T. [tex]\alpha[/tex] and solving the first order condition reveals that the R.H.S. is unconditionally maximised when

[tex]\alpha=\alpha^*=1-\frac{64 n (1-\gamma )+101 \gamma }{\gamma \left(128-27 \gamma ^n\right)}[/tex]

(the second derrivative is nagative, so that the R.H.S is concave). Now, note that

[tex]\frac{\partial\alpha^*}{\partial \gamma}=\frac{n \left(8192+27 (-64+64 n (-1+\gamma )-101 \gamma ) \gamma ^n\right)}{\gamma ^2 \left(128-27 \gamma ^n\right)^2}[/tex],

which has the same sign as [tex]\left(8192+27 (-64+64 n (-1+\gamma )-101 \gamma ) \gamma ^n\right)[/tex].

Moreover,

[tex]\frac{\partial}{\partial\gamma}\left(8192+27 (-64+64 n (-1+\gamma )-101 \gamma ) \gamma ^n\right)=-27 (1+n) (64 n (1-\gamma )+101 \gamma ) \gamma ^{n-1}\leq0[/tex].

This implies that [tex]\partial\alpha^*/\partial\gamma[/tex] can be negative for some [tex]\gamma\in(0,1)[/tex] only if it is negative for [tex]\gamma=1[/tex]. Making this substitution reveals that

[tex]\left.\frac{\partial\alpha^*}{\partial \gamma}\right|_{\gamma=1}=\frac{37n}{101}>0\ \Rightarrow\ \frac{\partial\alpha^*}{\partial \gamma}>0.[/tex]

Thus, the largest admissible value of [tex]\alpha^*[/tex] occurs where [tex]\gamma=1[/tex], in which case [tex]\left.\alpha^*\right|_{\gamma=1}=0[/tex]. Concavity and the fact that [tex]\alpha\in(0,1)[/tex] must be true then implies that the right hand side of the reduced inequality is maximised by its limit as [tex]\alpha[/tex] goes to zero. Taking the limit of LHS of the original inequality first as [tex]\delta[/tex] goes to zero and then as [tex]\alpha[/tex] goes to zero reduces that inequality to

[tex]
\frac{101 \gamma -128 n \gamma ^n+2 (-37+64 n) \gamma ^{1+n}-27 \gamma ^{1+2 n}}{512 \gamma }\geq0.
[/tex]

Multiplying away extraneous terms and simplifying then leaves

[tex]\gamma ^{n-1} \left(128 n (1-\gamma )+\gamma \left(74+27 \gamma ^n\right)\right)\leq101[/tex],

but I still don't know how to show that this holds over the relative range of parameter values.
 
Last edited:

Answers and Replies

  • #2
verty
Homework Helper
2,164
198
In what sense is this an ongoing research project? Is it a task you are supposed to complete on your own?
 
  • #3
To calrify: This is an ongoing research project in the sense that I am an independent researching academic (albeit not a proper mathematician) who has encountered the above problem in the course of an applied modelling project in my own academic research. I am hoping that some fresh eyes might spot a trick that I have missed and thus help in obtaining the proof, or at least suggest some alternative avenue of investigation.

It is because this is a part of original research that I can't be sure that a proof exists. However, plotting the reduced inequality, it seems pretty clear that the LHS is maximised at 101 (the function is smooth and continuous and approaches 101 at a corner solution). I am therefore pretty confident that the inequality does indeed hold. Alas, even though the reduced inequality appears to behave monotonically in both n and gamma, showing as much has also defied me.

I am not sure whether one of the main mathematics forums would have been a better place for this post, but I am not able to move the thread.
 
  • #4
I think I managed to finish off the proof. The remaining steps are shown in the attached image for anyone who is interested, or encounters a similar problem.
 

Attachments

  • #5
verty
Homework Helper
2,164
198
I think this works.

Let [tex]k = \gamma^n[/tex]

[tex]27 k^2 - (128n - 74) k - 101 <= - 128 n k^{1 - 1/n}[/tex]

1. Both sides are decreasing in k,
2. For k -> 0, LHS < RHS
3. d/dk LHS <= d/dk RHS
 
  • #6
Vertigo, that indeed appears to work and is far more elegant that the method I posted--which helps me with my page count. Many thanks.
 

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