Number of solutions via generating function How to find generating function for number of solutions
$x_1+x_2+x_3=n$ in set of positive integers such that $x_1 \ge x_2 \ge x_3$ and $x_1<x_2+x_3$. 

Consider the diagram

where the number of dots in column $1$ is no greater than the number of dots in column $2$, which is no greater than the number of dots in column $3$, which is equal to the number of dots in column $4$ (columns $3$ and $4$ are identical).
The diagram represents a solution to
$$
a\le b\le c\quad\text{and}\quad c\le a+b\quad\text{and}\quad a+b+c=n\tag1
$$
where $a$ is the number of dots in column $2$, $b$ is the number of dots in column $3$, $c$ is the sum of the number of dots in columns $1$ and $4$, and $n$ is the number of dots in all columns. All solutions to $(1)$ can be uniquely represented this way.
We can count the number of solutions to $(1)$ by looking at the diagrams sideways. Each diagram represents a term in the expansion of
$$
\frac1{1-x^2}\frac1{1-x^3}\frac1{1-x^4}\tag2
$$
The diagram above represents the $x^{16}$ term from
$$
\underbrace{\color{#AAA}{\overbrace{\left(1+\color{#000}{x^2}+x^4+x^6+\cdots\right)}^{\large\frac1{1-x^2}}}}_\text{$1$ rows with $2$ dots}
\underbrace{\color{#AAA}{\overbrace{\left(1+x^3+\color{#000}{x^6}+x^9+\cdots\right)}^{\large\frac1{1-x^3}}}}_\text{$2$ rows with $3$ dots}
\underbrace{\color{#AAA}{\overbrace{\left(1+x^4+\color{#000}{x^8}+x^{12}+\cdots\right)}^{\large\frac1{1-x^4}}}}_\text{$2$ rows with $4$ dots}\tag3
$$
The diagram

represents the $x^{16}$ term from
$$
\underbrace{\color{#AAA}{\overbrace{\left(1+x^2+\color{#000}{x^4}+x^6+\cdots\right)}^{\large\frac1{1-x^2}}}}_\text{$2$ rows with $2$ dots}
\underbrace{\color{#AAA}{\overbrace{\left(\color{#000}{1}+x^3+x^6+x^9+\cdots\right)}^{\large\frac1{1-x^3}}}}_\text{$0$ rows with $3$ dots}
\underbrace{\color{#AAA}{\overbrace{\left(1+x^4+x^8+\color{#000}{x^{12}}+\cdots\right)}^{\large\frac1{1-x^4}}}}_\text{$3$ rows with $4$ dots}\tag4
$$
Thus, the number of such diagrams with $n$ dots is
$$
\left[x^n\right]\frac1{1-x^2}\frac1{1-x^3}\frac1{1-x^4}\tag5
$$
That is, the generating function for the number of solutions to $(1)$ is given by $(2)$.

Each solution to $(1)$ can be mapped uniquely to a solution of
$$
a\le b\le c\quad\text{and}\quad c\lt a+b\quad\text{and}\quad a+b+c=n+3\tag6
$$
by adding $1$ to $a$, $b$, and $c$. Thus, the number of solutions to $(6)$ is given by $(5)$.
Therefore, the number of solutions to
$$
a\le b\le c\quad\text{and}\quad c\lt a+b\quad\text{and}\quad a+b+c=n\tag7
$$
is
$$
\left[x^{n-3}\right]\frac1{1-x^2}\frac1{1-x^3}\frac1{1-x^4}=\left[x^n\right]\frac{x}{1-x^2}\frac{x}{1-x^3}\frac{x}{1-x^4}\tag8
$$
Thus, the generating function for the number of solutions to $(7)$ is
$$
\bbox[5px,border:2px solid #C0A000]{\frac{x^3}{\left(1-x^2\right)\left(1-x^3\right)\left(1-x^4\right)}}\tag9
$$
