Prove an inequality on natural number Show that if $ a,b\in N$ and $a < b$, 
then $$\frac{a^a}{(a+1)^{a+1}} > \frac{b^b}{(b+1)^{b+1}}.$$

Using Bernoulli's Inequality, we get
$$
\begin{align}
\frac{(a+1)^{a+1}}{a^a}\frac{(a-1)^{a-1}}{a^a}
&=\frac{a+1}{a-1}\left(1-\frac1{a^2}\right)^a\\
&\ge\frac{a+1}{a-1}\left(1-\frac1a\right)\\
&=\frac{a+1}a\\[8pt]
&\gt1
\end{align}
$$
Therefore,
$$
\frac{a^a}{(a+1)^{a+1}}\lt\frac{(a-1)^{a-1}}{a^a}
$$
which says that $\dfrac{a^a}{(a+1)^{a+1}}$ is a strictly decreasing function of $a$.
