Inequality with floor and a fraction

Question

Given positive integers b and v satisfying the condition $1 \le v \lt b$, prove that

$$\left \lfloor{\frac{b}{v+1}}\right \rfloor \ge \frac{b-v}{v+1}$$

Any idea or a hint? I tried solving it starting with the definition of floor, but to no avail.

Answer 1

Since $b$ must be an integer, notice that $$\frac b{v+1}-\lfloor\frac b{v+1}\rfloor\leq\frac v{v+1}$$

as if this is not the case, $$\frac b{v+1}-\lfloor\frac b{v+1}\rfloor>\frac v{v+1}$$ Since $b$ is an integer, this implies$$\frac b{v+1}-\lfloor\frac b{v+1}\rfloor\geq\frac {v+1}{v+1}$$

which is not possible. This shows after rearranging that the inequality is satisfied.

Answer 2

Note that $\lfloor x\rfloor > x-1$ for any real number $x$. So $$\left\lfloor\frac b{v+1}\right\rfloor> \frac b{v+1} - 1 = \frac{b-v-1}{v+1}.$$

Answer 3

Let $b = \alpha(v+1) + \beta$ where $ 0 \le \beta < v+1$.

Note that $0 \le \beta \le v$; so $v-\beta \ge 0$.

Then $$\left\lfloor\dfrac b{v+1}\right\rfloor = \alpha$$

and

\begin{align} \frac{b-v}{v+1} &= \frac{\alpha(v+1) + \beta-v}{v+1} \\ &= \alpha - \frac{v-\beta}{v+1} \\ &\le \alpha \end{align}

Answer 4

The important fact is:

if $x$ and $y$ are integers and $x>y$, then $x\ge y+1$.

We have $$\Bigl\lfloor\frac b{v+1}\Bigr\rfloor>\frac b{v+1}-1\ ;$$ multiplying both sides by $v+1$ gives $$(v+1)\Bigl\lfloor\frac b{v+1}\Bigr\rfloor>b-v-1\ ;$$ both sides are integers so $$(v+1)\Bigl\lfloor\frac b{v+1}\Bigr\rfloor\ge b-v\ ;$$ and hence $$\Bigl\lfloor\frac b{v+1}\Bigr\rfloor\ge \frac{b-v}{v+1}\ .$$