Prove an inequality on natural number

Question

Show that if $ a,b\in N$ and $a < b$, then $$\frac{a^a}{(a+1)^{a+1}} > \frac{b^b}{(b+1)^{b+1}}.$$

Answer 1

Hint. You may just observe that, for $x >0$, the function given by $$f(x)=\log \left(\dfrac{x^x}{(x+1)^{x+1}} \right) $$ is decreasing. Indeed, we have $$ f(x)=-x \log\left(1+\frac 1x\right)-\log (x+1) $$ and $$ f'(x)=- \log\left(1+\frac 1x\right)<0, \qquad x>0. $$

Answer 2

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 $a$.