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 $\mathbb{R}$
 A: hmmmmm. I am not sure. The property you mention is not generally a property (in general) of either concave or convex functions, but actually of sub-additive functions. However, I am inclined to believe that the conjecture, or the book may be correct for the following reasoning. 
Technically,  (on the average, to abuse terminology) a property that would most likely be shared by concave functions over the positive reals, rather then convex functions. Unless its one of these sub-linear or super-linear functions which are some admixture of both (and almost, both,'convex and concave', or super-additive, and sub-additive) over $[0,1]$) 
Notice that in $(1)$ the scope of the multiplicative elements $a$ and $(1-a)$ is within $F$ not outside $F$ as in $(2)$ and $(3)$. If one can show that $F$ is sub-additive or that F(tx)>=tF(x) it should be.
$(1)f(ax+(1−a)y)≤f(ax)+f((1−a)y)$
Convex: $f(ax+(1−a)y) ≤a f(x)+(1−a)f(y)$
Concave:  $f(ax+(1−a)y)\geq af(x)+(1−a)f(y)$
I agree that $(1)$ definitely Seems wrong for concave $F$ but it actually may not be.  Remember that  the  "above" $(1)$ is a property that function, $F$ 'sub-additive over the positive reals' may have.
Concave functions are  sub-additive,  over the positive reals, when $f(0) ≥ 0$,see https://en.wikipedia.org/wiki/Concave_function. See pt 10.
Although the domain only specifies positive- reals, not-non-negative reals and may not  include the $0$. That is the only the issue. It also not a closed domain and range either.
However, given the positive domain,  and strict monotonic increasing-ness, it might have the same effect, and the conjecture MIGHT (I stress) be correct. It will also be strictly quasi convex and strictly quasi concave. 
If $F$ is concave and,$f(0) ≥ 0$ then $F$  is sub-additive over the positive reals; so the book, is "arguably" correct.
I  (stress) arguably, as the  the domain, of the function, only specifies positive- reals, not-negative reals. ie including $0$. Maybe strict mono-tonicity, and posi-tivity may help in this case. Notice, that given sub-additivity. 
A: 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 $\frac{a}{c}+\frac{b}{c}=1, \land\, (\frac{b}{c},\frac{a}{c})\,\in [0,1]$
