Let $f:[0,\infty) \to [0,\infty)$ be increasing and satisfy $f(0)=0$ and $f(x)>0 \forall x>0$. Let $f:[0,\infty) \to [0,\infty)$ be increasing and satisfy $f(0)=0$ and $f(x)>0 
\forall x>0$. If $f$ also satisfies $f(x+y) \leq f(x)+f(y) \forall x,y \geq 0$, then $f \circ d$ is a metric whenever $d$ is metric. Show each of the following conditions is sufficient to ensure that $f(x+y) \leq f(x)+f(y) \forall x,y \geq 0$:
a) $f$ has a second derivative satisfying $f'' \leq 0$;
b) $f$ has a decreasing first derivative.
c) $f(x)/x$ is decreasing for $x>0$.
My Work Thus Far:
Let $g(x)=f(x)x$, for $x>0$. According to the question we have that $f''<0$ then $f$ has a decreasing first derivative, meaning $f'$ is decreasing. Now, taking the derivative of $g$ yields: $g'(x)=xf'(x)-f(x)x^2=f'(x)x-f(x)x^2$, where $x>0$.
I know that $-f(x)x^2$ is decreasing, but how do I know for sure that $f'(x)x$ is decreasing so I can deduce that $g'(x)<0, \forall x>0$ and that $g(x)$ is a decreasing function?

By $c$ you have $f(x+y)\cdot(x+y)\leq f(x)\cdot x$ and $f(x+y)\cdot(x+y)\leq f(y)\cdot y$. Equivalently: $$f(x+y)\cdot\dfrac{x+y}x\leq f(x), \  f(x+y)\cdot\dfrac{x+y}{y}\leq f(y)$$
Summing both inequalities gives: $$f(x+y)\left(\dfrac{x+y}{x}+\dfrac{x+y}{y}\right)\leq f(x)+f(y)$$
From here you can conclude expanding and aplying AM-GM inequality.
