How to find/prove the maximum of a symmetric homogeneous function involving sides of a triangle? Let a, b and c be the side lengths of a triangle. Consider the following function 
$$ f(a,b,c)=\frac{\sqrt{b\,c\,(a+c-b)(a+b-c)}}{a\,(a+b+c)} + \frac{\sqrt{a\,c\,(b+c-a)(a+b-c)}}{b\,(a+b+c)} + \frac{\sqrt{a\,b\,(b+c-a)(a+c-b)}}{c\,(a+b+c)} $$.
It is a symmetric and homogeneous function with degree 0. With the help of the computer algebra system Maple, I learn that it has a maximum 1 when a=b=c, but I don't know how to prove that the maximum of f(a,b,c) is 1. Maple only provides the answer when I use the command "Maximize" within the package "Optimization". Could anyone provide an analytic(readable) proof for the conclusion? Thanks
One more observation: Since it is a homogeneous function with degree 0, f(a, b, c)=f(a.1, a.(b/a), a.(c/a))=f(1, b/a, c/a). So, WLOG, we can assume a=1. Consequently the function is reduced to f(1, b, c) with only two variables.

By AM-GM, we have $\sqrt{(a+c-b)(a+b-c)} \le \frac{a+c-b + a+b-c}{2} = a$
(or, $(a+c-b)(a+b-c) = a^2 - (b-c)^2 \le a^2$).
So, the first term is less than or equal to $\frac{\sqrt{bc}}{a+b+c}$.
Similarly, the second term is less than or equal to $\frac{\sqrt{ac}}{a+b+c}$,
and the third term is less than or equal to $\frac{\sqrt{ab}}{a+b+c}$.
So, $f(a,b,c) \le \frac{\sqrt{bc} + \sqrt{ac} + \sqrt{ab}}{a+b+c} \le 1$
(by letting $a = x^2, b = y^2, c = z^2$, it is just $x^2 + y^2 + z^2 \ge xy + yz + zx$
which is obvious).
