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:
$$1+2abcxyz \geq a^2x^2 + b^2y^2+c^2z^2$$
 A: We'll rewrite our conditions in the following forms.
$(1-b^2)(1-c^2)=(a-bc)^2$ and $(1-y^2)(1-z^2)\geq(x-yz)^2$.
Let $b^2=\frac{1}{1+v^2}$, $c^2=\frac{1}{1+w^2}$, $y^2=\frac{1}{1+q^2}$, $z^2=\frac{1}{1+r^2}$, $a-bc=ubc$ and $x-yz=pyz$,
where $v$, $w$, $q$ and $r$ are non-negatives.
Hence, our conditions give $vw\geq|u|$, $qr\geq|p|$ and we need to prove that
$$(1-b^2y^2)(1-c^2z^2)\geq(ax-bcyz)^2$$ or
$$\left(1-\frac{1}{(1+v^2)(1+q^2)}\right)\left(1-\frac{1}{(1+w^2)(1+r^2)}\right)\geq((1+u)(1+p)bcyz-bcyz)^2$$ or
$$((1+v^2)(1+q^2)-1)((1+w^2)(1+r^2)-1)\geq((1+u)(1+p)-1)^2$$ or
$$(v^2q^2+v^2+q^2)(w^2r^2+w^2+r^2)\geq(up+u+p)^2,$$
which is true by C-S:
$$(v^2q^2+v^2+q^2)(w^2r^2+w^2+r^2)\geq(vwqr+vw+qr)^2\geq$$
$$\geq\left(|up|+|u|+|p|\right)^2\geq(up+u+p)^2.$$
Done!
A: 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.
