How to prove this inequality $f(a+b)\leq f(a) +f(b)$ for $\frac{f(x)}x$ monotone decreasing

Question

$f(x)$ is defined on$(0,\infty)$, and $\frac{f(x)}{x}$ is monotone decreasing, how to prove $\forall a>0,b>0$ $f(a+b)\leq f(a) +f(b)$. Thanks pretty much in advance

Answer 1

Let $a>b>0$. Thus, $$f(a+b)<(a+b)\frac{f(a)}{a}=f(a)+\frac{f(a)}{a}\cdot b<f(a)+\frac{f(b)}{b}\cdot b=f(a)+f(b).$$

Answer 2

\begin{align*} &\dfrac{f(a)}{a}\geq\dfrac{f(a+b)}{a+b}\\ &\dfrac{f(b)}{b}\geq\dfrac{f(a+b)}{a+b}, \end{align*} so \begin{align*} &f(a)\geq\dfrac{a}{a+b}\cdot f(a+b)\\ &f(b)\geq\dfrac{b}{a+b}\cdot f(a+b), \end{align*} adding together, we have \begin{align*} f(a)+f(b)\geq\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}\right)f(a+b)=f(a+b). \end{align*}