<s>[INST] inequality:$(a+c)(b+d)(ac+bd)+4\ge 3(a+b+c+d)$ 
If $a,b,c,d>0$ such that $abcd=1$ prove that $$(a+c)(b+d)(ac+bd)+4\ge 3(a+b+c+d)$$

My first idea was to find counter examples and also check the bahaviour of inequality in the edge cases.Easy to see the equality occurs when $a=b=c=d=1$ However there is a big problem when $a\to 0^+ b\to 0^+,c\to 0^+,d\to \infty$ In this case the RHS $\to \infty$ and the LHS although appears $\to 0$ we see that as $b+d\to \infty$ so we cannot say much.
Mixed variables doesnt help much as we cannot take WLOG $a\ge b\ge c\ge d$
I also tried Holder $$(a+c)(b+d)(bd+ac)\ge {(\sqrt[3]{ab^2d}+\sqrt[3]{ac^2d})}^3=ad{(b^{2/3}+c^{2/3})}^3$$ which  has only complicated it further
Looking forward to solutions without lagranges multiplier.
 [/INST] WLOG assume that $bd\ge 1$ and $ac\le 1$. The inequality rewrites as
$$(a+c)((b+d)(ac+bd)-3)+4\ge 3(b+d).$$
AM-GM and $bd\ge 1$ yield
$$(b+d)(ac+bd)-3\ge 2\sqrt{bd}\cdot 2\sqrt{acbd}-3 \ge 2\cdot 2-3=1>0,$$
hence by AM-GM
$$(a+c)((b+d)(ac+bd)-3) \ge 2\sqrt{ac}((b+d)(ac+bd)-3) = 2(b+d)((ac)^{3/2}+(bd)^{1/2})-6(bd)^{-1/2}.$$
Hence it is enough to prove that
$$2(b+d)((ac)^{3/2}+(bd)^{1/2}-\frac 32) + 4 \ge 6(ac)^{1/2} .$$
Using AM-GM again we obtain $$(ac)^{3/2}+(bd)^{1/2} \ge 4\sqrt[4]{(ac)^{3/2}\cdot \left(\frac{(bd)^{1/2}}{3}\right)^3}=4\cdot 3^{-3/4}>\frac 32,$$
hence by AM-GM
$$2(b+d)((ac)^{3/2}+(bd)^{1/2}-\frac 32) \ge 4(bd)^{1/2}\left((ac)^{3/2}+(bd)^{1/2}-\frac 32\right) = 4ac+4bd-6(bd)^{1/2}$$
and we are left to prove that
$$4ac+4bd+4\ge 6(ac)^{1/2} + 6(bd)^{1/2}.$$
This follows from AM-GM:
$$4ac+4bd+4\ge 3ac+3bd+6 = 3ac+3 + 3bd+3 \ge 6\sqrt{ac}+6\sqrt{bd}$$
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