How to prove this inequality $\sqrt{\prod\limits_{cyc}(a+\sqrt{\frac{bcd}{a}})}+2\sqrt{abcd} \ge ab+bc+cd+da+ac+bd$ Let $a,b,c,d>0$. Show that
$$\sqrt{\left(a+\sqrt{\dfrac{bcd}{a}}\right)\left(b+\sqrt{\dfrac{acd}{b}}\right)\left(c+\sqrt{\dfrac{abd}{c}}\right)\left(d+\sqrt{\dfrac{abc}{d}}\right)}+2\sqrt{abcd}\ge ab+bc+cd+da+ac+bd$$
My idea: make use of
$$a+\sqrt{\dfrac{bcd}{a}}=a+\dfrac{\sqrt{abcd}}{a}\ge 2\sqrt{\sqrt{abcd}}$$
It is said this inequality can be proved using AM-GM( Cauchy-Schwarz) inequality. But I can't.
Thank you for you help in proving it.

WLOG, assume $abcd = 1$. The desired inequality is written as
$$\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)} + 2 \ge ab + bc + cd + da + ac + bd.$$
Since $ab + bc + cd + da + ac + bd \ge 6$, it suffices to prove that
$$(a^2+1)(b^2+1)(c^2+1)(d^2+1) - (ab + bc + cd + da + ac + bd - 2)^2 \ge 0.$$
Using $d = \frac{1}{abc}$, we have
$$\mathrm{LHS} = \frac{(ab - 1)^2(bc-1)^2(ca-1)^2}{a^2b^2c^2}.$$
We are done.
