<s>[INST] 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.
 [/INST] For $a=b=c=d=1$ we obtain a value $5$.
We'll prove that it's a minimal value.
Indeed, let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2,$ 
$abc+abd+acd+bcd=4w^3$ and $abcd=t^4$.
Thus, by AM-GM $v^4\geq t^4,$ $v^2=1$ and we need to prove that:
$$abcd+\sqrt{a^2b^2c^2d^2+\sum\limits_{cyc}a^2b^2c^2+\frac{1}{4}\sum\limits_{sym}a^2b^2+\sum\limits_{cyc}a^2+1}\geq5$$ or
$$t^4+\sqrt{t^8+(16w^6-12v^2t^4)v^2+(36v^4-32uw^3+2t^4)v^4+(16u^2-12v^2)v^6+v^8}\geq5v^4$$ or
$$t^8-10v^4t^4+16v^2w^6+25v^8-32uv^4w^3+16u^2v^6\geq25v^8-10v^4t^4+t^8$$ or
$$(uv^2-w^3)^2\geq0$$ and we are done!
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