Proving an Inequality Using Functions Let $n$ be a positive integer and $x>0$. Prove the following:
$$\dfrac{x^n}{3}\geq \dfrac{1}{x+2}+\dfrac{(3n+1)\ln(x)}{9}$$
So I approached the problem by considering
$$f(n)=\dfrac{x^n}{3}-\dfrac{1}{x+2}-\dfrac{(3n+1)\ln(x)}{9}$$
$$f'(n)=\dfrac{x^n \ln(x)}{3}-\dfrac{\ln(x)}{3}=\dfrac{(x^n-1)\ln(x)}{3}$$
I noticed that $f'(n)\geq 0$ for all $n\geq 1, \quad x>0$ but I couldn't make any meaningful observations from here.

Since $f(n)$ is an increasing function, it is enough to prove the original inequality for $n=1$, but
$$ \frac{1}{x+2}+\frac{4\log x}{9}\leq \frac{x}{3} \tag{1}$$
is a convexity inequality: the LHS of $(1)$, say $g(x)$, is a concave function, since
$$ \frac{d^2}{dx^2}\left(\frac{1}{x+2}+\frac{4\log x}{9}\right) = \frac{2}{(2+x)^3}-\frac{4}{9x^2}<0\tag{2}$$
is equivalent to $2(x+2)^3>9x^2$ or to $16 + 24 x + 3 x^2 + 2 x^3>0$, trivial for $x\in\mathbb{R}^+$, and the equation of the tangent line to $g(x)$ at $x=1$ is exactly $y=\frac{x}{3}$.
