Let $b$ be a fixed positive number. Define the following function for $x>0$:
$$ f(x) = \frac{(x+1)^{b+1}}{x^b} $$
Let us find the critical points of $f$ (the points where the derivative is zero). By the quotient rule the numerator of the derivative is equal to (we will ignore the denominator, as you will see why):
$$ \left[ (x+1)^{b+1} \right]' x^b - (x+1)^{b+1} \left[ x^b \right]' $$
This becomes,
$$ (b+1)(x+1)^b x^b - b(x+1)^{b+1}x^{b-1} $$
This is the numerator for $f'(x)$. To find the critical points of $f$ we just need to set the numerator equal to zero:
$$ (b+1)(x+1)^b x^b - b(x+1)^{b+1}x^{b-1} = 0$$
Divide both sides by $(x+1)^bx^{b-1}$ to end up with:
$$ (b+1)x - b(x+1) = 0 \implies bx + x - bx - b = 0 \implies x = b$$
Therefore, as $b$ is the critical point it means $f$ is either increasing or decreasing on the interval $(0,b)$ and that it is either increasing or decreasing on the interval $(b,\infty)$. Note that,
$$ \lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} \frac{(x+1)^{b+1}}{x^b} = \infty $$
This means that $f$ must be decreasing on $(0,b)$.
Also note that,
$$ \lim_{x\to \infty} f(x) = \lim_{x\to \infty} \frac{(x+1)^{b+1}}{x^b} = \infty $$
Therefore, $f$ must be increasing on $(b,\infty)$.
This means that the point $b$ is a global minimum point for $f$. Thus, we have that $f(x) \geq f(b)$ for any point $x$. Now choose $x=a$ where $a$ is any positive number and we get that,
$$ f(a) \geq f(b) \implies \frac{(a+1)^{b+1}}{a^b} \geq \frac{(b+1)^{b+1}}{b^b} \implies \frac{(a+1)^{b+1}}{(b+1)^{b+1}} \geq \frac{a^b}{b^b} \implies \left( \frac{a+1}{b+1} \right)^{b+1} \geq \left( \frac{a}{b} \right)^b $$
