Is $f $ decreasing?

Question

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.

Answer 1

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 $3f(x+y+z)\le f(x)+f(y)+f(z).$

Answer 2

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!