# Inequality in natural numbers

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?

• You've also gone backwards: you proved $B \Rightarrow A$ when you wanted $A \Rightarrow B$. Aug 7 '18 at 15:04
• You are right! So, anyway, my "proof" is completely incorrect. Aug 7 '18 at 15:10

$$\frac{v(b-v)}{v+1}-\frac{b-1}{2}=\frac{1}{2}\frac{(v-1)(b-1-2v)}{v+1}$$