23rd Junior Turkish Mathematical Olympiad

Answer: $c=1$. If $x=y=z=\sqrt[6]{4c}$ then we get $$\frac{x^{3}y+y^{3}z+z^{3}x}{x+y+z}+\frac{4c}{xyz}=4\sqrt{c}\ge 2c+2$$ and hence $(\sqrt{c}-1{)}^2\le 0$. Therefore, all $c\ne 1$ do not satisfy the inequality. Let us show that the inequality holds for $c=1$. By AM-GM inequality we have

$$x^{3}y+\frac{4}{xy}\ge 4x$$ $$y^{3}z+\frac{4}{yz}\ge 4y$$ $$z^{3}x+\frac{4}{zx}\ge 4z$$ Side by side summation of these inequalities yields $$x^{3}y+y^{3}z+z^{3}x+\frac{4(x+y+z)}{xyz}\ge 4(x+y+z).$$ Therefore, $$\frac{x^{3}y+y^{3}z+z^{3}x}{x+y+z}+\frac{4}{xyz}\ge 4$$ and we are done.