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## 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.

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