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.
Best Answer
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 $$1+y/2<1+y = (1+y)^1\leqslant (1+y)^y\leqslant (n+1+y)^y.$$