Showing 'simple' inequality

Question

I want to show two claims, which are closely connected, for $\delta \in (0,1)$, $n\in \mathbb{N}$.

1. $\frac{\delta-\delta^{n-1}}{1-\delta^n}<\frac{n-2}{2}$ for all $n>2$.
2. $ \frac{1-\delta^k}{k}\cdot \frac{k+1}{1-\delta^{k+1}} > \frac{1-\delta^b}{b}\cdot \frac{b+1}{1-\delta^{b+1}}$ for all $k<b$.

Both inequalities hold for $\delta=0$. Furhtermore, for $\delta \rightarrow 1$ the LHS of the first inequality becomes $\frac{n-2}{n}$. Similarly, for $\delta \rightarrow 1$ both sides of the second inequality approach 1.

For both inequalities a monotonicity arguemnt approach should work, but I didn't manage it. Thanks for your ideas!

Answer 1

I have found a way to prove the first inequality. Let $\delta \in (0,1)$ andnote that in general we have that $\delta^{n-1}>\delta^n$ for $n>2$. This means $1-\delta^{n-2} < 1-\delta^n$ and thus $\frac{1-\delta^{n-1}}{1-\delta^n}$ (easy to see that if the exponent difference is larger this also holds). Now we prove the first inequality!

Take $n=3$. Filling this in gives $$\frac{\delta - \delta^2}{1-\delta^3}= \frac{\delta(1-\delta)}{1-\delta^3}<\frac{\delta(1-\delta)}{1-\delta^2} =\frac{\delta}{1+\delta}<\frac{1}{2}=\frac{3-2}{2}$$ So for $n=3$ the inequality holds.

Notice that for $n\geq 4$ the solution is simpler. $$\frac{\delta - \delta^{n+1-1}}{1-\delta^{n+1}}=\frac{\delta(1-\delta^{n-1})}{1-\delta^{n+1}}< \delta <1 \leq\frac{n-2}{2}$$ Clearly $\frac{n-2}{2} \geq 1$ for $n>3$.

I have not tried my hand at the second inequality yet, but I imagine a similar method will do but you might have to use induction to prove it. Try if you can take it from here!

Answer 2

Disclaimer: For the first one you can prove much stronger result, namely $$ \frac{\delta - \delta^{n-1}}{1 - \delta^n} < \frac{n - 2}n. $$ Note that the desired result follows from the above inequality immediately, since $n > 2$.

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Since $1 - \delta^n > 0$ then, it's equivalent to proving $$ f(\delta) := (2 -n) \delta^n + n\delta^{n-1} - n\delta + n - 2 > 0, ~ \forall \delta \in (0, 1) $$ First order condition (FOC) gives: $$ f'(\delta) = n((2-n)\delta^{n-1} + (n-1) \delta^{n-2} - 1) = 0 $$ In can be easily shown that $f(\cdot)$ is convex by taking the second derivative and ensuring that it is indeed positive, without much details you should get $$ f''(\delta) = n(n - 1)(n - 2) \delta^{n-3}(1 - \delta) > 0, ~ \forall \delta \in (0, 1) $$ So, we know that the points satisfying FOC are the minimum points, which implies that it is sufficient to show the inequality for them only.

Note that $(2 - n)\delta^{n-1} = 1 - (n - 1)\delta^{n-2}$ from FOC, hence $$ f(\delta) = \delta - (n - 1)\delta^{n-1} + n\delta^{n-1} - n\delta + n - 2 = \delta(1- n) + n - 2 + \delta^{n-1} = \delta^{n-1} - 1 + (n-1)(1 - \delta) = (\delta - 1)(\delta^{n-2} + \dots + \delta + 1 - n + 1) > 0, $$ since both multipliers in the final expression are negative, which completes the proof.

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For the second one notice that it suffices to show that the function $$f(n) := \frac{1 - \delta^n}{n} \cdot \frac{n + 1}{1 - \delta^{n+1}}, ~ \forall n \in \mathbb{N}$$ is decreasing. It's true for all $x \in \mathbf{R}_+$, so we'll prove for this case. The tedious calculations and simplifications of the first derivative yield to the following expression (substituting $n$ with $x$) $$ (\delta^{x+1} - 1)^2 \left[\delta^{2x + 1} + \delta^x\left[((1 - \delta)x(x + 1) + \delta + 1)) + 1\right]\right] > 0, $$ which is indeed true.