The Problems of Ukrainian Authors

$abc=1\Rightarrow \frac{1}{3}(a+b+c)$. Then: $$\begin{array}{rl}{c}^2+a+b& \le c^2+(a+b)\frac{1}{3}(a+b+c)\\ & =\frac{1}{3}(3c^2+a^2+b^2+2ab+ac+bc)\\ & \le \frac{1}{6}(8c^2+5a^2+5b^2).\end{array}$$ So $$\frac{a^2+b^2}{c^2+a+b}+\cdots \ge \frac{a^2+b^2}{\frac{1}{6}(8c^2+5a^2+5b^2)}+\cdots +\frac{6(a^2+b^2)}{8c^2+5a^2+5b^2}+\cdots \ge 2,$$ the last inequality must be proved. As three points meant a cyclic permutation of variables. For simplification denote $x=a^2$, $y=b^2$, $z=c^2$. Then we need to prove such inequality: $$A=\frac{x+y}{8z+5x+5y}+\cdots \ge \frac{1}{3}.$$ $$A=\frac{(x+y{)}^2}{(x+y)(8z+5x+5y)}+\cdots \ge \frac{((x+y{)}^2+\dots {)}^2}{(x+y)(8z+5x+5y)}=\frac{4p+8q}{10p+26q}\ge \frac{1}{3},$$ this inequality supervenes from Cauchy-Schwarz inequality. There are used next denote: $p=x^2+y^2+z^2$, $q=xy+yz+zx$. The last inequality equivalent $p\ge q$, which are well known.