For integer $x \ge 1$, does it follow that $\left(\frac{x^2+2x+1}{x^2+2x}\right)^{x+1} > \frac{x+2}{x+1}$ Would I be correct that the answer here is yes?
Here is my thinking:

*

*For $x \ge 1$, $\dfrac{x+2}{x+1}$ is strictly decreasing.

*For $x \ge 1$, $\left(\dfrac{x^2 + 2x + 1}{x^2 + 2x}\right)^{x+1}$ is strictly increasing

*$\dfrac{16}{9} > \dfrac{3}{2}$
To show that $\left(\dfrac{x^2 + 2x + 1}{x^2 + 2x}\right)^{x+1}$ is strictly increasing I take the derivative which is positive for $x > 0$

As commenters have pointed out, the left side is not increasing, but decreasing.
You can see this by showing:
$$\left(\frac{x^2+2x+1}{x^2+2x}\right)^{(x+1)^2}\to e,$$ so the left side of your inequality is in the range of $e^{1/(x+1)},$ and converges to $1.$

Bernoulli's inequality says that if $y\geq -1$ and $n$ is a non-negative integer, then:
$$(1+y)^n\geq 1+ny\tag1$$
We only need it for $y> 0,$ where $(1)$ obviously follows from the binomial theorem.
In your case, $n=x+1, y=\frac{1}{x^2+2x}$ gives you:
$$\left(\frac{x^2+2x+1}{x^2+2x}\right)^{x+1}\geq 1+\frac{x+1}{x^2+2x}>1+\frac{x+1}{x^2+2x+1}=1+\frac1{x+1}=\frac{x+2}{x+1}$$
