, 2015

Let $d = D(a, b)$. We can write $a = a_1 d$ and $b = b_1 d$, where $a_1$ and $b_1$ are co-prime. We have $v(a, b) = a_1 b_1 d$. We insert this in the condition of the problem to get
$$a_1 d + b_1 d \mid d + a_1 b_1 d \Rightarrow a_1 + b_1 \mid 1 + a_1 b_1,$$
and we can rearrange the inequality to
$$\frac{d + a_1 b_1 d}{a_1 d + b_1 d} \le \frac{a_1 d + b_1 d}{4} \Leftrightarrow 1 + a_1 b_1 \le d \frac{(a_1 + b_1)^2}{4} \Leftrightarrow 4 \le d(a_1^2 + b_1^2) + (2d - 4)a_1 b_1.$$
If $d \ge 2$ we have
$$d(a_1^2 + b_1^2) + (2d - 4)a_1 b_1 \ge 2(a_1^2 + b_1^2) + (2 \cdot 2 - 4)a_1 b_1 = 2(a_1^2 + b_1^2) \ge 2(1^2 + 1^2) = 4,$$
Lets now check the case when $d = 1$. We must prove the inequality
$$4 \le a_1^2 + b_1^2 - 2a_1 b_1 \Leftrightarrow 4 \le (a_1 - b_1)^2 \Leftrightarrow 2 \le |a_1 - b_1|.$$
This means we have to prove that the numbers $a_1$ and $b_1$ differ by at least 2. Hence it suffices to check if that they do not differ for less.
If $a_1 = b_1$ we get $a_1 = b_1 = 1$ due to co-primality and hence $a = b = 1$, which is impossible.

If the numbers differ by 1 we may assume that $b_1 = a_1 + 1$ since the other case is symmetric. We insert this to the condition of the problem to get
$$\begin{align*} a_1 + (a_1 + 1) \mid 1 + a_1(a_1 + 1) & = a_1^2 + a_1 + 1 \\ \Rightarrow \quad 2a_1 + 1 \mid 2(a_1^2 + a_1 + 1) & = (2a_1 + 1)a_1 + a_1 + 2 \\ \Rightarrow \quad 2a_1 + 1 \mid a_1 + 2 \end{align*}$$
However, since $2a_1 + 1 \ge a_1 + 2$ this is possible only if $a_1 = 1$. Since $d = 1$ we again get $a = 1$, which is impossible since $a, b > 1$.
Thus we proved that the numbers $a_1$ and $b_1$ differ by at least 2 which proves the inequality.