Let $\frac{a+c}{2\sqrt{ac}}=x$ and $\frac{b+d}{2\sqrt{bd}}=y$. Hence, $x\geq1$, $y\geq1$ and we need to prove that $x^2y^2-x-y+1\geq0$, which is $xy(xy-1)+(x-1)(y-1)\geq0$, which is obvious.

http://www.artofproblemsolving.com/community/c6h1091287p4853924

$\blacklozenge \color{red}{\textit{\textbf{Proof:}}}$ We have \begin{align*} abcd+16 &=ab+bc+cd+da+16\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{cd}+\frac{1}{da}\right) \\ &=ab+\frac{16}{bc}+cd+\frac{16}{da}+bc+\frac{16}{cd}+da+\frac{16}{ab} \\ &\ge 8\left(\sqrt \frac{a}{c}+\sqrt \frac{c}{a} \right)+8\left(\sqrt \frac{b}{d} +\sqrt \frac{d}{b} \right) \\ &=8\left(\sqrt {\frac{a}{c}+2+\frac{c}{a}}\right)+8\left(\sqrt {\frac{b}{d}+2+\frac{d}{b}}\right) \\ &=8 \sqrt{(a+c)\left(\frac{1}{a} + \frac{1}{c}\right)}+8\sqrt{(b+d)\left(\frac{1}{b}+\frac{1}{d}\right)}. \quad \blacksquare \end{align*}

The condition is equivalent to $(a+c)(b+d)=cd+da+ab+bc=abcd$. By AM-GM, \begin{align*} abcd+16&\geq 8\sqrt{abcd}=8\sqrt{(a+c)(b+d)}. \end{align*}Now we need, \begin{align*} \sqrt{(a+c)(b+d)}&\geq \sqrt{(a+c)\left(\frac{bd}{b+d}\right)}+\sqrt{(b+d)\left(\frac{ac}{a+c}\right)}\Longleftrightarrow\\ 2(a+c)(b+d)&\geq 2(a+c)\sqrt{bd}+2(b+d)\sqrt{ac}, \end{align*}which is true as $(a+c)(b+d)\geq 2(a+c)\sqrt{bd}$ and $(a+c)(b+d)\geq 2(b+d)\sqrt{ac}$ by AM-GM. The equality holds iff $a=c,b=d$ and $ab=4$.

Pretty easy for an Iran TST $RHS=\frac{8(a+c)\sqrt{bd}+8(b+d)\sqrt{ac}}{\sqrt{(a+c)(b+d)}}$ $\le$ $2\sqrt{16(a+c)(b+d)}$ $\le$ $16+(a+c)(b+d)=16+abcd=LHS$ .