Three. (20 points) Given the function $f(x)$ for any real numbers $x, y$ has
$$
f(x+y)=f(x)+f(y)+2(x+y)+1,
$$

and $f(1)=1$.
If $x \in \mathbf{N}_{+}$ and $x \geqslant 2$, the inequality
$$
f(x) \geqslant(a+7) x-(a+10)
$$

always holds, try to find the range of the real number $a$.
Let $y=1$, then
$$
f(x+1)=f(x)+f(1)+2(x+1)+1,
$$

which means $f(x+1)-f(x)=2 x+4$.
When $x \in \mathbf{N}_{+}$, we have
$$
\begin{array}{l}
f(2)-f(1)=2 \times 1+4, \\
f(3)-f(2)=2 \times 2+4, \\
\cdots \cdots \\
f(x)-f(x-1)=2(x-1)+4 .
\end{array}
$$

Adding the above equations, we get
$$
f(x)-f(1)=2 \times \frac{1+(x-1)}{2}(x-1)+4(x-1).
$$

Simplifying, we get $f(x)=x^{2}+3 x-3$.
When $x \in \mathbf{N}_{+}$ and $x \geqslant 2$,
$f(x) \geqslant(a+7) x-(a+10)$ always holds
$\Leftrightarrow x^{2}+3 x-3 \geqslant(a+7) x-(a+10)$ always holds
$\Leftrightarrow x^{2}-4 x+7 \geqslant a(x-1)$ always holds.
Since $x \geqslant 2$, we have $\frac{x^{2}-4 x+7}{x-1} \geqslant a$ always holds.
$$
\begin{array}{l}
\text { Also, } \frac{x^{2}-4 x+7}{x-1}=\frac{(x-1)^{2}-2(x-1)+4}{x-1} \\
=(x-1)+\frac{4}{x-1}-2 \geqslant 2 \sqrt{(x-1) \frac{4}{x-1}}-2=2 .
\end{array}
$$

The equality holds if and only if $x-1=\frac{4}{x-1}$, i.e., $x=3$.
Therefore, the minimum value of $\frac{x^{2}-4 x+7}{x-1}$ is 2.
Thus, the range of real number $a$ is $(-\infty, 2]$.