Problems from Ukrainian Authors

We are given the following inequalities: $$1+2abc\ge a^2+b^2+c^2\text{ and }1+2xyz\ge x^2+y^2+z^2,(1)$$ and we need to show that the following inequality holds: $$1+2abcxyz\ge (ax{)}^2+(by{)}^2+(cz{)}^2.(2)$$ 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: $$abxy-\sqrt{(1-a^2{x}^2)(1-b^2{y}^2)}\le cz\le abxy+\sqrt{(1-a^2{x}^2)(1-b^2{y}^2)}.$$ 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 $$(ab+\sqrt{(1-a^2)(1-b^2)})(xy+\sqrt{(1-x^2)(1-y^2)})\le abxy+\sqrt{(1-a^2{x}^2)(1-b^2{y}^2)}.(3)$$ 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 $$a\sqrt{(1-x^2)(1-y^2)}\le \sqrt{(1-a^2{x}^2)(1-y^2)},$$ which is now obvious since $a\le 1,1-a^2{x}^2\ge 1-x^2$. 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.