The Problems of Ukrainian Authors

Let's use $u$, $v$, $w$ - method. You can get more information about this method in numerous articles, e.g. ($x=p-a$, $y=p-b$, $z=p-c$): $$R=\frac{abc}{4S}=\frac{(x+y)(y+z)(z+x)}{4\sqrt{xyz}(x+y+z)}\text{and}r=\frac{S}{p}=\sqrt{\frac{xyz}{x+y+z}}$$

Then we have to prove equivalent inequality: $$(x+y)(y+z)(z+x)+2(3\sqrt{3}-4)xyz\ge 2(x+y+z)\sqrt{xyz}(x+y+z).$$ Let's denote: $$\{\begin{array}{l}x+y+z=3u,\\ xy+yz+zx=3v^2,\text{Then }9u{v}^2+(6\sqrt{3}-9)w^3\ge 6\sqrt{3}\sqrt{u{w}^3}u\\ xyz=w^3.\end{array}$$

$$\begin{gathered} (3u{v}^2+(2\sqrt{3}-3)w^3{)}^2\ge 12u^3{w}^3\iff \\ f(w^3)=(2\sqrt{3}-3{)}^2{w}^6+w^3(6(2\sqrt{3}-3)u{v}^2-12u^3)+9u^2{v}^4\ge 0. \end{gathered}$$

If $u$, $v^2$ are fixed then $f$ is convex. That's why $f$ has a minimum value when 2 variables $\{x,y,z\}$ coincide. Without loss of generality $y=z$. Let's check our inequality. It's equivalent to the following: $$(x+y)((2x+y)-xy)+2(3\sqrt{3}-4)xy\ge 2\sqrt{x(x+2y{)}^3}.$$ Let's denote: $t=\frac{x}{y}\Rightarrow$ $$\begin{gathered} (t+2)(t+1)+2(3\sqrt{3}-4)t\ge 2\sqrt{t(t+2{)}^3}\iff t^2+3(3\sqrt{3}-2)t+1\ge \sqrt{t(t+2{)}^3}\iff \\ {t}^3(6\sqrt{3}-10)+t^2(21-12\sqrt{3})+t(6\sqrt{3}-12)+1\ge 0\iff (t-1{)}^2((6\sqrt{3}-10)t+1)\ge 0. \end{gathered}$$ What was to be demonstrated. The equation is true for $x=y=z\iff a=b=c$.