Is $f $ decreasing? Let $f:(0,\infty)\rightarrow (0,\infty) $, $f (x)+f (y)\geq 2f (x+y), \forall x, y>0$.
Is $f $ decreasing?
I need to show that $f (x)+f (y)+f (z)\geq 3f (x+y+z) $ and it's enough to show that $f$ is decreasing.
 A: Here is an alternative way to prove your property.
Note that $4f(x+y+z)\le2f(x)+2f(y+z)\le2f(x)+f(y)+f(z)$.
By symmetry $4f(x+y+z)\le2f(y)+f(x)+f(z)$ and $4f(x+y+z)\le2f(z)+f(x)+f(y)$.
Summing up the inequalities give $12f(x+y+z)\le4f(x)+4f(y)+4f(z)$,
dividing by four gives $3f(x+y+z)\le f(x)+f(y)+f(z).$
A: We have:
$$f(x)+f(y)+f(z) \ge 2f(x+y)+f(z) = f(x+y) +f(z) + f(x+y)+f(z)-f(z) \ge 2f(x+y+z)+2f(x+y+z)-f(z)=4f(x+y+z)-f(z)$$
In the same manner:
$$f(x)+f(y)+f(z) \ge 2f(x+z)+f(y) = f(x+z) +f(y) + f(x+z)+f(y)-f(y) \ge 2f(x+y+z)+2f(x+y+z)-f(y)=4f(x+y+z)-f(y)$$
And:
$$f(x)+f(y)+f(z) \ge 2f(y+z)+f(x) = f(y+z) +f(x) + f(y+z)+f(x)-f(x) \ge 2f(x+y+z)+2f(x+y+z)-f(x)=4f(x+y+z)-f(x)  $$
In the end we have:
$$ f(x)+f(y)+f(z) \ge 4f(x+y+z)-f(z)$$
$$ f(x)+f(y)+f(z) \ge 4f(x+y+z)-f(y)$$
$$ f(x)+f(y)+f(z) \ge 4f(x+y+z)-f(x)$$
Summing the we get:
$$3(f(x)+f(y)+f(z)) \ge 12f(x+y+z)-f(x)-f(y)-f(z) \Leftrightarrow 4(f(x)+f(y)+f(z)) \ge 12f(x+y+z) \Leftrightarrow f(x)+f(y)+f(z) \ge 3f(x+y+z)$$
I wish I helped!
