, 2023

The answer is 6. The equality holds when
$$(a, b, c, d) = \left( \sqrt[4]{6}, \sqrt[4]{\frac{27}{2}}, \frac{\sqrt{6}}{2}, 1 \right).$$
Solution 1. By Cauchy-Schwarz inequality we get
$$\left[ (a^2 + 3d^2) + (b^2 + 2c^2) \right] \left[ \frac{a^2}{a^2 + 3d^2} + \frac{b^2}{b^2 + 2c^2} \right] \geq (a+b)^2$$
and
$$\left[ (2a^2 + 6d^2) + (3b^2 + 6c^2) \right] \left[ \frac{3d^2}{a^2 + 3d^2} + \frac{2c^2}{b^2 + 2c^2} \right] \geq 6(c+d)^2$$
Then it follows that
$$S := \frac{(a+b)^2}{a^2 + b^2 + 2c^2 + 3d^2} + \frac{6(c+d)^2}{2a^2 + 3b^2 + 6c^2 + 6d^2} \\ \le \frac{a^2}{a^2 + 3d^2} + \frac{b^2}{b^2 + 2c^2} + \frac{3d^2}{a^2 + 3d^2} + \frac{2c^2}{b^2 + 2c^2} = 2$$
Finally by Arithmetic-Geometric Mean inequality, we have
$$\frac{6(a+b)^2(c+d)^2}{(a^2+b^2+2c^2+3d^2)(2a^2+3b^2+6c^2+6d^2)} \le \frac{S^2}{4} \le 1$$
and hence we are done.

Solution 2. It follows from Cauchy-Schwarz inequality that
$$(a^2 + b^2 + 2c^2 + 3d^2)(6c^2 + 6d^2 + 3b^2 + 2a^2)$$
$$\geq 6(ad + bc + ac + bd)^2 = 6(a+b)^2(c+d)^2$$
and hence the minimal value of the expression is 6.