For postive real numbers $a$,$b$,$c$,$d$ and $e$, prove that
$$\frac{4a^3}{a^2+2b^2+\frac{2b^3}{a}} + \frac{4b^3}{b^2+2c^2+\frac{2c^3}{b}} + \frac{4c^3}{c^2+2d^2+\frac{2d^3}{c}}+ \frac{4d^3}{d^2+2e^2+\frac{2e^3}{d}} + \frac{4e^3}{e^2+2a^2+\frac{2a^3}{e}} \geq $$ $$ \frac{2ab^2+2b^3}{a^2+2b^2 + 2\frac{b^3}{a} } + \frac{2bc^2+2c^3}{b^2+2c^2 + 2\frac{c^3}{b} }+ \frac{2cd^2+2d^3}{c^2+2d^2 + 2\frac{d^3}{c} } + \frac{2de^2+2e^3}{d^2+2e^2 + 2\frac{e^3}{d} } + \frac{2ea^2+2a^3}{e^2+2a^2 + 2\frac{a^3}{e} }$$
Can this be solved directly by $ AM-GM $?
By AM-GM: $$\sum_{cyc}\frac{4a^3}{a^2+2b^2+\frac{2b^3}{a}}-\sum_{cyc}\frac{2ab^2+2b^3}{a^2+2b^2+\frac{2b^3}{a}}=2\sum_{cyc}\frac{2a^4-a^2b^2-ab^3}{a^3+2ab^2+2b^3}=$$ $$=2\sum_{cyc}\left(\frac{2a^4-a^2b^2-ab^3}{a^3+2ab^2+2b^3}-(a-b)\right)=2\sum_{cyc}\frac{a^4+a^3b-3a^2b^2-ab^3+2b^4}{a^3+2ab^2+2b^3}=$$ $$=2\sum_{cyc}\frac{a^4-3a^2b^2+2ab^3+a^3b-3ab^3+2b^4}{a^3+2ab^2+2b^3}=2\sum_{cyc}\frac{(a+b)(a^3-3ab^2+2b^3)}{a^3+2ab^2+2b^3}\geq$$ $$\geq2\sum_{cyc}\frac{(a+b)\left(3\sqrt[3]{a^3\cdot\left(b^3\right)^2}-3ab^2\right)}{a^3+2ab^2+2b^3}=0.$$