I recently stumbled over a problem I am not able to solve. I want to show that
$$ (a+1-b)\cdot(a+1)^{b-1}\leq a^b,\quad a,b \in \mathbb{N}_{++}, a\geq b $$
According to Wolfram Alpha this seems to be true, however I am failing to prove this.
What I tried to do is to take the log and use some inequality on the log ($\log(x+1)\leq x$). However. this bound seems to be too wide. I get
$$ ab-b \leq b\log a $$ but from there I'm not able to find an upper bound for $b\log a$ that satisfies the inequality.
Does someone have some hints on other inequalities to look into? (No homework, real problem). Thanks!
$$\iff \frac{a+1-b}{a+1}\leq \left(\frac{a}{a+1} \right)^b\iff 1-\frac{b}{a+1} \leq \left(1-\frac{1}{a+1} \right)^b$$ Let's study the function $f(x)=(1-x)^b+bx-1$ for $b\in\mathbb{N}_{++}$ and $x>0$.
We have $f'(x)=b(1-(1-x)^{b-1})\geq 0$ for $x>0$ and $b \geq 1$.
Then $f(x)$ is increasing function and $f(x) \geq f(0) = 0$. (Q.E.D)