Proving $\frac{a}{b^3}+\frac{b}{c^3}+\frac{c}{a^3}\geqslant \frac{a+b}{b^3+c^3}+\frac{b+c}{c^3+a^3}+\frac{c+a}{a^3+b^3}$

We need to prove that: $$\sum_{cyc}(a^{10}c^6+a^9b^7-a^7b^6c^3-a^6b^6c^4)\geq0,$$ which is true by AM-GM:

$$\sum_{cyc}a^{10}c^6=\frac{1}{38}\sum_{cyc}\left(14a^{10}c^6+21b^{10}a^6+3c^{10}b^6\right)\geq$$ $$\geq\sum_{cyc}\sqrt[38]{a^{14\cdot10+21\cdot6}b^{21\cdot10+3\cdot6}c^{16\cdot6+3\cdot10}}=\sum_{cyc}a^7b^6c^3$$ and $$\sum_{cyc}a^9b^7=\frac{1}{67}\sum_{cyc}\left(33a^9b^7+19b^9c^7+15c^9a^7\right)\geq$$ $$\geq\sum_{cyc}\sqrt[67]{a^{33\cdot9+15\cdot7}b^{33\cdot7+19\cdot9}c^{19\cdot7+15\cdot9}}=\sum_{cyc}a^6b^6c^4$$ and we are done!