strictly increasing concave function on R+ the real analysis book says that 
$$f:\mathbb{R}_+ \rightarrow \mathbb{R}_+$$ where $f$ is strictly increasing and concave function. it has the following property
$$f(ax+(1-a)y) \le f(ax) + f((1-a)y)$$
where $a \in [0,1]$.
This property seems wrong. As far as I know, that property is for convex function, not concave function. I do not think the textbook is wrong. Can you please explain it?
The textbook used it to show the function $d(x,y)$ is a matric in $\mathbb{R}$

See also:https://en.wikipedia.org/wiki/Subadditivity:,  which gives, Remember that these are sufficient conditions.
Notice that given sub-additivity:
\begin{align}
F\left(\frac{a}{a+b}\times (a+b) + \frac{b}{a+b} \times (a+b)\right)
& = F(a + b) \\
& \le F(a) + F(b) \\
& = F\left(\frac{a}{a+b}\times  (a+b)\right) + F\left(\frac{b}{a+b}\times  (a+b)\right),
\end{align}
where presumably $\frac{b}{a+b}$ $\in (0,1)$ if $a$,$b$ and $a+b$ are positive real numbers, as per the domain and range.
If, the function above is sub-additive over the positive reals.
$$F\left(\frac{a}{c} (c) + \frac{b}{c}(c)\right)
\le F\left(\frac{a}{c}(c)\right) + F\left(\frac{b}{c}(c)\right)$$
Where $\frac{a}{c}+\frac{b}{c}=1, \land\, (\frac{b}{c},\frac{a}{c})\,\in [0,1]$
