10. Let $f$ be a real function defined on all positive integers, $f(1)=\frac{1}{2}$ and
$$
f(n+1) \geqslant \frac{3 f(n)}{2 f(n)+1}(n \geqslant 1),
$$

try to find $f(n)$.
10. From the known conditions, it is easy to see that $f(n)>0$. Taking the reciprocal on both sides of (1), we get
$$
2+\frac{1}{f(n)} \geqslant \frac{3}{f(n+1)}(n \geqslant 1) \text {. }
$$

Subtracting 3 from both sides, we get
$$
\frac{1}{f(n)}-1 \geqslant 3\left[\frac{1}{f(n+1)}-1\right] \text {, }
$$

Multiplying both sides by $3^n$, we get
$$
3^{n}\left[\frac{1}{f(n)}-1\right] \geqslant 3^{n-1}\left[\frac{1}{f(n+1)}-1\right],
$$

Let $3^{n}\left[\frac{1}{f(n)}-1\right]=k(n)$. Satisfying $g(1)=3\left[\frac{1}{f(1)}-1\right]=3 \cdot g(n) \geqslant g(n+1)$,
Thus, the solution to the given functional inequality is
$$
f(n)=\frac{3^{n}}{3^{n}+g(n)},
$$

where $g(n)$ is any function satisfying $g(1)=3 \cdot g(n) \geqslant g(n+1)\left(n \in \mathbf{N}^{+}\right)$.