, 2017

By the Cauchy-Schwarz inequality, we have
$$(a^2b + c^2d + ad^2 + cb^2)(a^2d + c^2b + ab^2 + cd^2) \geq (a^2\sqrt{bd} + c^2\sqrt{bd} + abd + CBD)^2 = \left(\frac{a^2 + c^2}{\sqrt{ac}} + abd + CBD\right)^2.$$
Together with $a^2 + c^2 \geq \frac{1}{2}(a+c)^2 \geq (a+c)\sqrt{ac}$, we have
$$(a^2b + b^2c + c^2d + d^2a)(ab^2 + bc^2 + cd^2 + da^2) \geq (a+c)^2(1+bd)^2.$$
Due to symmetry, we also have
$$(a^2b + b^2c + c^2d + d^2a)(ab^2 + bc^2 + cd^2 + da^2) \geq (b+d)^2(1+ac)^2.$$
Multiplying these inequalities, we obtain
$$(a^2b + b^2c + c^2d + d^2a)(ab^2 + bc^2 + cd^2 + da^2) \geq (a+c)(b+d)(1+ac)(1+bd)$$
where $(1+ac)(1+bd) = 1+ac+bd+abcd = ac+bd+2$. The result follows readily.
For equality in the application of the Cauchy-Schwarz inequality, we need $b = d$ and $a = c$. Note that the other inequalities also have these as equality. Together with $abcd = 1$, equality holds when $a = c = t$ and $b = d = \frac{1}{t}$ for some $t > 0$. $\square$