Mathematical Reflections 2012, Issue 6 - Problem J248

Problem:
Let $f:[1,\infty) \longrightarrow \mathbb{R}$ be defined by $f(x)=\dfrac{\{x\}^2}{\lfloor x \rfloor}$. Prove that $f(x+y) \leq f(x)+f(y)$, for any real numbers $x$ and $y$.

Proposed by Sorin Radulescu.

Solution:
Since $\{x+y\} \leq \{x\}+\{y\}$ and $\lfloor x \rfloor + \lfloor y \rfloor \leq \lfloor x+y \rfloor$ for all $x,y \in \mathbb{R}$, using the well known inequality $$\dfrac{a^2}{x}+\dfrac{b^2}{y} \geq \dfrac{(a+b)^2}{x+y} \qquad \forall a,b,x,y \in \mathbb{R}, x,y>0$$ we have $$f(x+y)=\dfrac{\{x+y\}^2}{\lfloor x+y \rfloor} \leq \dfrac{(\{x\}+\{y\})^2}{\lfloor x \rfloor + \lfloor y \rfloor} \leq \dfrac{\{x\}^2}{\lfloor x \rfloor} + \dfrac{\{y\}^2}{\lfloor y \rfloor}=f(x)+f(y)$$ for all $x,y \in [1,\infty)$.