Proving $\frac{a}{b^3}+\frac{b}{c^3}+\frac{c}{a^3}\geqslant \frac{a+b}{b^3+c^3}+\frac{b+c}{c^3+a^3}+\frac{c+a}{a^3+b^3}$ For $a,b,c>0.$ Prove$:$ $$\dfrac{a}{b^3}+\dfrac{b}{c^3}+\dfrac{c}{a^3}\geqslant \dfrac{a+b}{b^3+c^3}+\dfrac{b+c}{c^3+a^3}+\dfrac{c+a}{a^3+b^3}\quad (\text{Tran Quoc Thinh}) $$
It's easy with Buffalo Way and computer so I will not post it.
(Please don't post solution by Buffalo Way, thanks for a real a lot!)
So$,$ we try to find a solution by hand.
I get this SOS$:$
$$\sum {\dfrac { \left( 6\,{a}^{5}+5\,{a}^{4}b+2\,{a}^{4}c+4\,{a}^{3}{b}^{2}+4
\,{a}^{3}bc+8\,{a}^{2}{b}^{3}+6\,{a}^{2}{b}^{2}c+3\,a{b}^{4}+4\,a{b}^{
3}c-2\,{b}^{5}+2\,{b}^{4}c \right)  \left( a-b \right) ^{2}}{{a}^{3}{b
}^{3} \left( a+b \right)  \left( {a}^{2}-ab+{b}^{2} \right) }} \geqslant 0,$$
By SOS theorem$,$ if $$S_a+S_b+S_c \geqslant 0 ; S_a S_b +S_b S_c +S_cS_a\geqslant 0.$$
Then $$S_a (b-c)^2 +S_b (c-a)^2 +S_c(a-b)^2\geqslant 0.$$
Here$,$ we can prove$:$ $$S_a+S_b+S_c \geqslant 0,$$
but $$S_a S_b +S_b S_c +S_cS_a\geqslant 0$$ is not true!
pqr or $uvw$ technique give a very high degree, I think it is impossble.

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!
