4
$\begingroup$

Consider the following inequality: \begin{gather} c\cdot(n+1)^n<(c-y)\cdot(n+1+y)^{n+y} \end{gather} where $n,c,y$ are all integers satisfying $n\geqslant0$, $c\geqslant3$, $y\geqslant1$ and $c-y\geqslant2$.

I have spent the last weeks trying to see whether the inequality above is true or not. I have tried to find suitable values of $n$, $c$ and $y$ that make it false, but I have found none. I have also tried to prove it is true by trying to find a simpler and larger LHS and / or a simpler and smaller RHS, but I haven’t been successful.

The goal of this question is to see whether someone can provide a proof strategy for me to verify it myself and, failing that, whether someone with access to powerful analytical software can find suitable values that make the inequality false.

Thank you all.

$\endgroup$
5
  • 1
    $\begingroup$ Here, is $n$ in intended to be an integer and $c,y$ are real numbers? $\endgroup$
    – user1266745
    Commented Jan 11 at 21:37
  • $\begingroup$ Actually, $n$, $c$ and $y$ are all integers. $\endgroup$
    – EoDmnFOr3q
    Commented Jan 11 at 21:41
  • 1
    $\begingroup$ The worst case for $c$ is $c=y+2$, so you can replace $c$ by $y+2$. $\endgroup$ Commented Jan 11 at 21:51
  • 1
    $\begingroup$ Induction might do it. Re arranging the inequality miht be a good idea. $\endgroup$ Commented Jan 11 at 21:52
  • 1
    $\begingroup$ Note that $c\ge y+2$ implies $c \ge 3$. $\endgroup$ Commented Jan 11 at 21:55

3 Answers 3

3
$\begingroup$

Let $$\theta = \frac{y}{c-y}>0,$$ so that $$c=y+\frac{y}{\theta}.$$ Note that since $$y+\frac{y}{\theta}=c\geqslant y+2,$$ $$y/\theta\geqslant 2,$$ and $$0<\theta \leqslant y/2.$$ Using $c=y+y/\theta$ and $c-y= y/\theta$, we rewrite the inequality as $$\Bigl(1+\frac{1}{\theta}\Bigr)y (n+1)^n < (y/\theta)(n+1+y)^{n+y},\tag{W}$$ which is equivalent to the desired inequality.

Divide both sides by $y$ and multiply both sides by $\theta$. We get $$(\theta+1)(n+1)^n < (n+1+y)^{n+y},$$ subject to the constraint that $0<\theta\leqslant y/2$. Obviously it suffices to prove the inequality for $\theta=y/2$. So we now have $$(1+y/2)(n+1)^n < (n+1+y)^{n+y}=(n+1+y)^n(n+1+y)^y.$$

Expand $$(n+1+y)^n = \sum_{k=0}^n \binom{n}{k}(n+1)^ky^{n-k},$$ which includes as the $k=n$ term the value $(n+1)^n$. So $(n+1+y)^n\geqslant(n+1)^n$, and it suffices to prove $$(1+y/2)(n+1)^n < (n+1)^n (n+1+y)^y.\tag{E}$$ Divide through by $(n+1)^n$ and note that $E$ reduces to $$1+y/2 < (n+1+y)^y.$$ But this is clearly true, since $$1+y/2<1+y = (1+y)^1\leqslant (1+y)^y\leqslant (n+1+y)^y.$$

$\endgroup$
2
$\begingroup$

Rewrite the relation as \begin{equation} y (n+1+y)^{n+y} < ((n+1+y)^{n+y} - (n+1)^n) c \end{equation} we see that if the result is true for the smallest possible $c$, it is true for all $c$. The smallest possible $c$ is $y+2$. It remains to prove that \begin{equation} 0 < f(y) := 2 (n+1+y)^{n+y} - (n+1)^n (y + 2) \end{equation} We have \begin{align} f'(y) &=2 \ln(n+1+y) (n+1+y)^{n+y} + 2(n+1+y)^{n+y-1} - (n+1)^n\\ &\ge 2 \ln(2) (n+1+y)^{n+y} - (n+1)^n\\ &\ge (n+1+y)^{n+y} - (n+1)^n\\ &\ge 0 \end{align} So $f$ increases with $y$, and \begin{align} f(1)&=2(n+2)^{n+1} - 3(n+1)^n\\ &> (n+1)^n (2(n+2) - 3)\\ &\ge (n+1)^n\\ & > 0 \end{align} so that $f(y)>0$ holds for all $y\ge 1$.

$\endgroup$
2
  • $\begingroup$ Thank you very much for your answer. However, why does $0<f(n)$ not matter? $\endgroup$
    – EoDmnFOr3q
    Commented Jan 11 at 22:39
  • 1
    $\begingroup$ @EoDmnFOr3q I don't understand your question. We need to prove that $0<f(y)$. It is true that $f$ also depends on $n$ but I treat it as a fixed non negative parameter. $\endgroup$ Commented Jan 12 at 6:11
1
$\begingroup$

Similar to jwhite's answer, but shorter. Note that the case $n=0$ is trivial by binomial expansion, and using $c-y \ge 2$. So, for $n\ge 1$: $$\begin{aligned} \frac{c}{c-y} &= 1 + \frac{y}{c-y} \\ &\le 1+\frac{y}2 \\ &\le 1+\frac{n}{n+1}y \\ &\le \left(1+\frac{y}{n+1}\right)^n \\ &< \left(\frac{n+1+y}{n+1}\right)^n(n+1+y)^y \end{aligned}$$ Cross multiplying gives the desired inequality.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .