Vtyjkyuk

Let v and b be natural numbers satisfying the condition: $1 le v lt fracb2$. Prove that
$$fracv(b-v)v+1 ge fracb-12$$
(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 $v$:

$$2v^2-(b+1)v+b-1 le 0$$

$$ Delta = left(b-3right)^2$$

$$ sqrtDelta = lvert b - 3 rvert $$

Let's assume $b ge 3$ for a moment. Then we have:

$$v_1 = 1$$
$$v_2 = fracb-12$$

and we get the solution:

$$1 le v le fracb-12$$

Since b is natural, it's the same as:

$$1 le v lt fracb2$$

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

$$fracv(b-v)v+1-fracb-12=frac12frac(v-1)(b-1-2v)v+1$$