For a concave function, $f(\cdot)$ and that $f(0)\geq 0$, taking inputs $>0$, then we have the following relation $f(a)+f(b)\geq f(a+b)$. Assume also that $f(x)\geq f(y)$ for any $x>y$.
Does this imply that the following relation holds $(a,b,c>0)$; $$f(a+b)+f(a+c)\geq f(a+b+c)+f(a) $$I tried the following: $$f(a+b)+f(a+c)=f\left((a+b+c)\frac{a+b}{a+b+c}\right)+f\left((a+b+c)\frac{a+c}{a+b+c}\right)$$ Then, using concavity: $$\geq\frac{a+b}{a+b+c}f(a+b+c)+\frac{a+c}{a+b+c}f(a+b+c).$$ But dont know where if this is correct, or where if I can continue from here?
No we do not, not even with your extra condition that $f$ is nondecreasing. Take $f(x) = x-1$, $a=1$ and $b=0$. Then $f(a)+f(b) = -1$ while $f(a+b)=0$.