How to find the minimum of $abcd+\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}$ when $ab+bc+cd+da+ac+bd=6$ If $a,b,c,d\geq0$ are such that $ab+bc+cd+da+ac+bd=6$, then what is the minimum value of $$f(a,b,c,d)=abcd+\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}$$ ?
My attempts: I think the minimum $5$ is obtained at $a=b=c=d=1$. I tried the Hölder inequality: $$f(a,b,c,d)\geq abcd + (1+\sqrt{a b c d})^2,$$ but the last term is $1<5$ for $d=0$.
Also, direct AM-GM on the two terms of $f$ did not work either. What to do?
Remark: This problem is from AoPS. There, it is also asked what happens for $ab+bc+cd+da+ac+bd=7$, which seems even harder.

Using the known identity $(u^2 + v^2)(x^2 + y^2) = (ux - vy)^2 + (uy + vx)^2$,
we have
$$(a^2 + 1)(b^2 + 1) = (ab - 1)^2 + (a + b)^2 \tag{1}$$
and
$$(c^2+1)(d^2+1) = (cd - 1)^2 + (c + d)^2. \tag{2}$$
From (1) and (2), using Cauchy-Bunyakovsky-Schwarz inequality, we have
\begin{align*}
 &(a^2 + 1)(b^2 + 1)(c^2+1)(d^2+1)\\
 =\,& [(ab - 1)^2 + (a + b)^2]\,[(cd - 1)^2 + (c + d)^2]\\
 \ge\,& [(ab - 1)(cd - 1) - (a + b)(c + d)]^2\\
 =\,& (abcd - ab - cd + 1 - ac - ad - bc - bd)^2
\end{align*}
which results in
$$\sqrt{(a^2 + 1)(b^2 + 1)(c^2+1)(d^2+1)}
\ge ab + bc + cd + da + ac + bd - 1 - abcd.$$
Thus, we have
$$abcd + \sqrt{(a^2 + 1)(b^2 + 1)(c^2+1)(d^2+1)}
\ge ab + bc + cd + da + ac + bd - 1 = 5.$$
Also, when $a = b = c = d = 1$,
we have $abcd + \sqrt{(a^2 + 1)(b^2 + 1)(c^2+1)(d^2+1)} = 5$.
Thus, the minimum of $abcd + \sqrt{(a^2 + 1)(b^2 + 1)(c^2+1)(d^2+1)}$ is $5$.
