All-Soviet-Union Mathematical Olympiad

Solution:

Answer: $\sqrt{2} - 1 / 2$.

Obviously $a + d = b + c$ at the minimum value, because increasing $a$ or $d$ reduces the value. So we may take $d = b + c - a$. We also take $b \gg c$ (interchanging $b$ and $c$ if necessary). Dividing through by $b / 2$ shows that there is no loss of generality in taking $b = 2$, so $0 < c \leq 2$. Thus we have to find the minimum value of $2 / (2c - a + 2) + c / (a + 2)$. We show that it is $\sqrt{2} - 1 / 2$.

This is surprisingly awkward. Note first that $(c - (h - k))^{2} \geq 0$, so $c^{2} + c(2k - 2h) + h^{2} - 2hk + k^{2} \geq 0$. Hence $c^{2} + ck + h^{2} \geq (2h - k)(c + k)$. Hence $c / h^{2} + 1 / (c + k) \geq 2 / h - k / h^{2}$ with equality iff $c = h - k$. Applying this to $c / (a + 2) + 1 / (c + 1 - a / 2)$ where $h = \sqrt{a + 2}$, $k = 1 - a / 2$, we find that $c / (a + 2) + 2 / (2c + 2 - a) \geq 2 / \sqrt{a + 2} + (a - 2) / (2a + 4)$.

The allowed range for $c$ is $0 \leq c \leq 2$ and $0 \leq a \leq c + 2$, hence $0 \leq a \leq 4$. Put $x = 1 / \sqrt{a + 2}$, so $1 / \sqrt{6} \leq x \leq 1 / \sqrt{2}$. Then $2 / \sqrt{a + 2} + (a - 2) / (2a + 4) = 2x + 1 / 2 - 2x^{2} = 1 - (2x - 1)^{2} / 2$. We have $-0.184 = (2 / \sqrt{6} - 1) \leq 2x - 1 \leq \sqrt{2} - 1 = 0.414$. Hence $c / (a + 2) + 2 / (2c + 2 - a) \geq 1 - (\sqrt{2} - 1)^{2} / 2 = \sqrt{2} - 1 / 2$.

We can easily check that the minimum is achieved at $b = 2$, $c = \sqrt{2} - 1$, $a = 0$, $d = \sqrt{2} + 1$.