<s>[INST] Prove this $1+2abcxyz \geq a^2x^2 + b^2y^2+c^2z^2$ Let  $a,b,c,x,y,z\in\mathbb{R}$ such that \begin{align}\{a,b,c,x,y,z\}&\subset[-1,1]\\
1 + 2abc &\geq a^2+b^2+c^2\\
1+2xyz&\geq x^2+y^2+z^2\end{align}
Prove that:
$$1+2abcxyz \geq a^2x^2 + b^2y^2+c^2z^2$$
 [/INST] Let $A=\left(\begin{matrix}1 & a & b \\ a & 1 & c \\ b & c & 1 \end{matrix}\right)$. Due to assumptions on $a,b,c$, all principal minors of $A$ are nonnegative. Therefore $A$ is positive-semidefinite. 
Analogously, $B = \left(\begin{matrix}1 & x & y \\ x & 1 & z \\ y & z & 1 \end{matrix}\right)$ is positive-semidefinite. 
Let $C$ be the Hadamard product of $A$ and $B$, i.e. $$C = A \circ B = \left(\begin{matrix}1 & ax & by \\ ax & 1 & cz \\ by & cz & 1 \end{matrix}\right).$$ It follows from Schur product theorem that $C$ is positive-semidefinite as well. In particular $\det(C) \ge 0$, which is precisely the inequality we wanted to prove.
</s>