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).
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