Classical approach

Relevant wiki: Cauchy-Schwarz Inequality

Using Cauchy-Schwarz we get $\displaystyle\sum_{cyc}^{a,b,c}\frac{a^3+b^3}{a+2b}\geq \displaystyle\sum_{cyc}^{a,b,c}\frac{(a^2+b^2)^2}{(a+2b)(a+b)}$ By Titu's Lemma we have $\displaystyle\sum_{cyc}^{a,b,c}\frac{(a^2+b^2)^2}{(a+2b)(a+b)}\geq\frac{4(a^2+b^2+c^2)^2}{\displaystyle\sum_{cyc}^{a,b,c}(a+b)(a+2b)}=\frac{4(a^2+b^2+c^2)^2}{3(a^2+b^2+c^2+ab+ac+bc)}$ $\therefore\displaystyle\sum_{cyc}^{a,b,c}\frac{a^3+b^3}{a+2b}\geq\frac{4(a^2+b^2+c^2)^2}{3(a^2+b^2+c^2+ab+ac+bc)}$ Based on this inequality $ab+bc+ca\stackrel{C-S}\leq a^2+b^2+c^2$ $\Rightarrow\displaystyle\sum_{cyc}^{a,b,c}\frac{a^3+b^3}{a+2b}\geq\frac{4(a^2+b^2+c^2)^2}{3(a^2+b^2+c^2+ab+ac+bc)}\geq\frac{2}{3}(a^2+b^2+c^2)=2$ So the minimum value is 2 and the equality obtains when $a=b=c=1$