How to prove that $x\lfloor\frac{a}{x+1}\rfloor\geq \lfloor\frac{a}{2}\rfloor$, where $a\geq2$, $1\leq x\leq a-1$， $a$ and $x$ are integers.

Question

I have worked on this question for an hour. It since obvious if without $\lfloor\rfloor$, can I have a hint?

Answer 1

Since $ a \ge x+1$ we can let $a = m(x+1) + b$ where $0 \le b \le x$.

Then $$x\left \lfloor \frac{a}{x+1} \right \rfloor = mx = \left \lfloor \frac{2mx+1}{2} \right \rfloor \ge \left \lfloor \frac{m}{2}(x+1) + \frac x2 \right \rfloor \ge \left \lfloor \frac{m}{2}(x+1) + \frac b2 \right \rfloor = \left \lfloor \frac a2 \right \rfloor $$