It is just $\frac{a^2+b^2}{2} \leq a^2-ab+b^2 \leq a^2+b^2$

Left inequality: $\frac{a^2+b^2}{2} \leq 2 \cdot (a^3 + b^3) = 2 \cdot (a+b)(a^2-ab+b^2) = 2 \cdot(a^2-ab+b^2)$ is equivalent to $a^2 + b^2 \leq 2a^2-2ab+2b^2 \iff (a-b)^2 \geq 0$ which clearly is true. Now, the right inequality: $a^3 + b^3 \leq a^2 + b^2$ follows directly from $(a^2-ab+b^2) = (a+b)(a^2-ab+b^2) = a^3 + b^3 \leq a^2 + b^2 \iff ab \geq 0$ which is true as $a,b > 0$.

$\frac{a^3+(1-a)^3}{a^2+(1-a)^2}=r\iff \frac{3a^2-3a+1}{2a^2-2a+1}=r\iff (2r-3)a^2-(2r-3)a+(r-1)=0\implies a=\frac{2r-3\pm\sqrt{1-4(r-1)^2}}{(2r-3)^2}$ $1-4(r-1)^2\geq 0\implies(r-1)^2\leq\frac{1}{4}\implies -\frac{1}{2}\leq r-1\leq\frac{1}{2}\implies\frac{1}{2}\leq r\leq\frac{3}{2}$ However: $a,b\in(0,1]\implies a^3\leq a^2\wedge b^3\leq b^2\implies a^3+b^3\leq a^2+b^2\implies r\leq 1$ So the conclusion is $\frac{1}{2}\leq r\leq 1$.