Some Algebra Problem} 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 $n\geq 1,  x>0,$ but I was running
into some issues from there. In particular, given that
$$x^n ln(x) - ln(x) \geq 0 \implies x^n \geq 1,$$
I wonder what this is saying when $x < 1.$ Is it possible that I should be
handling this problem inductively?
 A: Induction seems like a good route. I believe you can avoid considering the derivative or what was answered by $\bf{Noah}$  $\bf{Schweber}$. See if you can prove that $$\dfrac{x}{3}\geq \dfrac{1}{x+2}+\dfrac{4\ln(x)}{9}$$ for all $x$ (base case of $n=1$). Without calculus it may be a bit algebra intensive but it's definitely doable. The rest is easier; we may now assume that $$\dfrac{x^k}{3}\geq \dfrac{1}{x+2}+\dfrac{(3k+1)\ln(x)}{9}$$ holds for some $k \geq 1$. It is clear that $$\begin{align}\dfrac{x^k}{3} +\frac{3\ln(x)}{9} &\geq \dfrac{1}{x+2}+\dfrac{(3k+1)\ln(x)}{9} +\frac{3\ln(x)}{9} \\ &= \dfrac{1}{x+2}+\dfrac{(3(k+1)+1)\ln(x)}{9} \end{align}$$ This puts the RHS of the inequality in the form we want. What remains to be shown is that $$\frac{x^{k+1}}{3} \geq \dfrac{x^k}{3} +\frac{3\ln(x)}{9}$$ Can you proceed from here? It may be helpful to use $\frac{x}{e} \geq \ln(x)$ for all $x$. Another thing that you may find useful is $$x^{k+1}-1 = x^k+x^{k-1}+\ldots + x+1$$
A: It is not true that $x^n\ln(x)-\ln(x)\ge 0$ implies $x^n\ge 1$.
To see this, note that $$x^n\ln(x)-\ln(x)=(x^n-1)\ln(x).$$ This is nonnegative if either $x^n-1$ and $\ln(x)$ are both nonnegative or both negative.
The former case happens exactly when both $x^n\ge 1$ and $x\ge 1$. But this is redundant: this is just $x\ge 1$.
If $x<1$, then $x^n-1$ is negative - but what is the sign of $\ln(x)$ for $0<x<1$?
