Let v and b be natural numbers satisfying the condition: $1 \le v \lt \frac{b}{2}$. Prove that $$\frac{v(b-v)}{v+1} \ge \frac{b-1}{2}$$ (remember we need that to be true only for natural $v$ and $b$)
How to prove it?
I started simply by multiplying both sides by $2(v+1)$, which gives a quadratic inequality after simple transformations with regards to $v$:
$$2v^2-(b+1)v+b-1 \le 0$$
$$ \Delta = \left(b-3\right)^2$$
$$ \sqrt{\Delta} = \lvert b - 3 \rvert $$
Let's assume $b \ge 3$ for a moment. Then we have:
$$v_1 = 1$$ $$v_2 = \frac{b-1}{2}$$
and we get the solution:
$$1 \le v \le \frac{b-1}{2}$$
Since b is natural, it's the same as:
$$1 \le v \lt \frac{b}{2}$$
which is our condition. Of course, we should also consider what happens when $b \lt 3$.
Anyway, I kind of think there's got to be a simpler way of proving it, not involving quadratic equations etc., just simple transformations making use of the fact that we want the inequality to be true only for natural v and b. Can someone show me a simpler reasoning leading to a simpler and more elegant proof?
Hint: Write your inequality in the form
$$\frac{v(b-v)}{v+1}-\frac{b-1}{2}=\frac{1}{2}\frac{(v-1)(b-1-2v)}{v+1}$$