Show $ f(x) = \sum_{n=1}^{\infty} \frac{nx}{n^3 + x^3}$ ,$\ g(x) = \sum_{n=1}^{\infty} \frac{x^4n}{(n^3 + x^3)^2}$ are bounded on $[0, \infty)$. If $f(x), g(x)$ are defined as following on $[0 , \infty)$,
$$\tag 1 f(x) = \sum_{n=1}^{\infty} \frac{nx}{n^3 + x^3}$$ 
$$\tag 2 g(x) = \sum_{n=1}^{\infty} \frac{x^4n}{(n^3 + x^3)^2}.$$
. Then how to show that $f,g$ are bounded function on $[0 , \infty)$?
I found this problem on this 
Is $f(x)=\sum_{n=1}^\infty\frac{nx^2}{n^3+x^3}$ uniformly continuous on $[0,\infty)$?.
In solution's last steps, i don't know why this is correct answer.
Could you explain for me to elaborate?
Thank you.

A simple proof:
Clearly:
$$
n^3+x^3\ge \frac{1}{4}(n+x)^3 \tag{1}
$$
and hence
$$
0\le \frac{nx}{n^3+x^3} \le \frac{4nx}{(n+x)^3}\le\frac{4(n+x)x}{(n+x)^3}=\frac{4x}{(n+x)^2}.
$$
Hence
$$
\sum_{n=1}^\infty\frac{nx}{n^3+x^3}\le 
\sum_{n=1}^\infty\frac{4x}{(n+x)^2}=\frac{4x}{(1+x)^2}+\sum_{n=2}^\infty\frac{4x}{(n+x)^2}\le \frac{4x}{(1+x)^2}+\int_1^\infty \frac{4x\,ds}{(s+x)^2} \tag{2}\\=
\frac{4x}{(1+x)^2}+\frac{4x}{(1+x)}.
$$
Meanwhile
$$
\frac{x^4n}{(x^3+n^3)^2}=\frac{nx}{n^3+x^3}\cdot\frac{x^3}{x^3+n^3}\le 
\frac{nx}{n^3+x^3}.
$$
Notes. Inequality $(1)$ holds since
$$
4(n^3+x^3)=4(n+x)(n^2-nx+x^2)=(n+x)(4n^2-4nx+4x^2)
=(n+x)(n^2+2nx+x^2+3n^2-6nx+3x^2)=(n+x)^3+3(n+x)(n-x)^2\ge(n+x)^3.
$$
The last inequality in $(2)$ holds since if $f: [1,\infty)\to (0,\infty)$ is decreasing, then $\sum_{n=2}^\infty f(n)\le \int_1^\infty f(x)\,dx.$
