Help with proving/disproving an inequality

Question

$\textbf{Question:}$Let $x_1,x_2,x_3,x_4 \in \mathbb{R}$ such that $(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1) =16 $. Is it then true that $x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4- x_1x_2x_3x_4 \le 5$, with equality $\iff x_1=x_2=x_3=x_4=\pm 1$?

Rough calculations seem to suggest that this is indeed the case, but I am unable to prove it.

For some context, this question is actually related to USAMO $2014$ P$1$. The original question was that: given a polynomial $P(x)=x^4+ax^3+bx^2+cx+d,$ and $ b-d \ge 5$, where all $4$ roots $x_1,x_2,x_3,x_4$ of $P(x)$ are real, find the smallest value of the expression $(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)$.

Indeed, I managed to prove that this expression is at least $16$. But to show that the minimum value of $16$ is actually attainable, I have to find a construction of some polynomial $P(x)$ satisfying the conditions of the question. While it is indeed obvious to see that setting $ x_1=x_2=x_3=x_4=1$ or $ x_1=x_2=x_3=x_4=- 1$ both works ( just expand $(x-1)^4$ and $(x+1)^4$ respectively), are there other non-trivial values that would work too?

In particular, using $\textbf{Vieta's formula}$ gives us that $b=x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4$ and $d=x_1x_2x_3x_4$, so the complicated looking expression in the first paragraph is actually just equivalent to $b-d$.

Answer 1

Since $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2,$$ we obtain: $$16=\prod_{k=1}^4(1+x_k^2)=((1-x_1x_2)^2+(x_1+x_2)^2)((1-x_3x_4)^2+(x_3+x_4)^2)=$$ $$=((x_1+x_2)(x_3+x_4)-(1-x_1x_2)(1-x_3x_4))^2+$$ $$+((1-x_1x_2)(x_3+x_4)+(1-x_3x_4)(x_1+x_2))^2\geq$$ $$\geq((x_1+x_2)(x_3+x_4)-(1-x_1x_2)(1-x_3x_4))^2=$$ $$=(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4- x_1x_2x_3x_4 -1)^2.$$ Can you end it now?

The equality occurs for $$(1-x_1x_2)(x_3+x_4)+(1-x_3x_4)(x_1+x_2)=0,$$ which is $$x_1+x_2+x_3+x_4=x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4$$ and for $$x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4- x_1x_2x_3x_4 -1=4.$$ We have for example also the following case of the equality occurring: $$(x_1,x_2,x_3,x_4)=\left(5,\frac{-5+\sqrt5}{2},\frac{-5-\sqrt5}{2},0\right).$$

Answer 2

Using the Cauchy-Schwarz inequality, we have $$(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)$$ $$= [(x_1+x_2)^2+(1-x_1x_2)^2][(x_3+x_4)^2+(x_3x_4-1)^2]$$ $$\geqslant [(x_1+x_2)(x_3+x_4)+(1-x_1x_2)(x_3x_4-1)]^2$$ $$=(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4-x_1x_2x_3x_4-1)^2.$$ Thefore $$(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4-x_1x_2x_3x_4-1)^2 \leqslant 16,$$ or $$x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4- x_1x_2x_3x_4 \leqslant 5.$$