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.
Best Answer
We have the following identity:
$$\prod_{cyc} (a^2+1) = \left(\sum_{cyc}a-\sum_{cyc}abc\right)^2+\left(\sum_{sym}ab-abcd-1\right)^2$$
So
$$\prod_{cyc} (a^2+1) \geq \left(\sum_{sym}ab-abcd-1\right)^2$$
and this implies
$$1+abcd+\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)} \geq ab+bc+cd+da+bd+ca$$
Now, we can see that the minimum is indeed $5$.