Problems from Ukrainian Authors

We are given the following inequalities:
$$1 + 2abc \ge a^2 + b^2 + c^2 \text{ and } 1 + 2xyz \ge x^2 + y^2 + z^2, \quad (1)$$
and we need to show that the following inequality holds:
$$1 + 2abcxyz \ge (ax)^2 + (by)^2 + (cz)^2. \quad (2)$$
First observation: we can assume that all numbers are non-negative. Indeed, if $abcxyz \ge 0$, then after taking the absolute values the inequality (2) does not change, while the inequalities (1) can only become stronger. If $abcxyz < 0$, then either $abc < 0$ or $xyz < 0$; without loss of generality, assume the former, then
$$1 + 2abcxyz \ge 1 + 2abc \ge a^2 + b^2 + c^2 \ge (ax)^2 + (by)^2 + (cz)^2.$$
Next, we can assume without loss of generality that $a \le b \le c$. Inequalities (1) and the left-hand side of (2) are invariant to the permutation of $x, y, z$. From the inequality for ordered tuples we get that the right-hand side of (2) is maximized when $x \le y \le z$, hence, it is enough to prove the statement only for this case.

Solving the inequality (2) as quadratic w.r.t. $cz$, we get the equivalent inequality:
$$abxy - \sqrt{(1 - a^2x^2)(1 - b^2y^2)} \le cz \le abxy + \sqrt{(1 - a^2x^2)(1 - b^2y^2)}.$$
The inequality on the left is obvious since $cz \ge ax \ge abxy$. Let us prove the inequality on the right. By solving the inequalities (1) as quadratic w.r.t. $c$ and $z$, we get
$$c \le ab + \sqrt{(1 - a^2)(1 - b^2)} \text{ and } z \le xy + \sqrt{(1 - x^2)(1 - y^2)}.$$
Thus, it is enough to show that
$$(ab + \sqrt{(1 - a^2)(1 - b^2)}) (xy + \sqrt{(1 - x^2)(1 - y^2)}) \le abxy + \sqrt{(1 - a^2x^2)(1 - b^2y^2)}. \quad (3)$$
Suppose that some of the numbers $a, b, x, y$ are equal to 1. Taking into account that $a \le b$ and $x \le y$, it is enough to consider the case $b=1$ (the case $y=1$ is similar). The inequality (3) becomes
$$a\sqrt{(1 - x^2)(1 - y^2)} \le \sqrt{(1 - a^2x^2)(1 - y^2)},$$
which is now obvious since $a \le 1, 1 - a^2x^2 \ge 1 - x^2$.
Now, suppose that none of the numbers is equal to 1. By expanding the brackets in the left-hand side and dividing by $\sqrt{(1 - a^2)(1 - b^2)(1 - x^2)(1 - y^2)}$, we get after minor simplifications the equivalent inequality:
$$1 + AB + XY \le \sqrt{(1 + A^2 + X^2)(1 + B^2 + Y^2)},$$
where $A = \frac{a}{\sqrt{1-a^2}}, B = \frac{b}{\sqrt{1-b^2}}, X = \frac{x}{\sqrt{1-x^2}}, Y = \frac{y}{\sqrt{1-y^2}}$. And the last inequality is a partial case of the Cauchy-Schwarz inequality.