<s>[INST] 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.
 [/INST] Partial Answer :
I rewrite the inequality like this with $a\geq c \geq b>0$:
$$\dfrac{a}{b^3}\Big(\frac{1}{2}-\frac{1}{\frac{c^3}{b^3}+1}\Big)+\dfrac{b}{c^3}\Big(\frac{1}{2}-\frac{1}{\frac{a^3}{c^3}+1}\Big)+\dfrac{c}{a^3}\Big(\frac{1}{2}-\frac{1}{\frac{b^3}{a^3}+1}\Big)\geq-\Big(\dfrac{b}{c^3}\Big(\frac{1}{2}-\frac{1}{\frac{b^3}{c^3}+1}\Big)+\dfrac{c}{a^3}\Big(\frac{1}{2}-\frac{1}{\frac{c^3}{a^3}+1}\Big)+\dfrac{a}{b^3}\Big(\frac{1}{2}-\frac{1}{\frac{a^3}{b^3}+1}\Big)\Big)$$
Or :
$$\dfrac{a}{b^3}\Big(\frac{1}{2}-\frac{1}{\frac{c^3}{b^3}+1}\Big)+\dfrac{b}{c^3}\Big(\frac{1}{2}-\frac{1}{\frac{a^3}{c^3}+1}\Big)+\dfrac{c}{a^3}\Big(\frac{1}{2}-\frac{1}{\frac{b^3}{a^3}+1}\Big)\geq\Big(\dfrac{b}{c^3}\Big(\frac{1}{2}-\frac{1}{\frac{c^3}{b^3}+1}\Big)+\dfrac{c}{a^3}\Big(\frac{1}{2}-\frac{1}{\frac{a^3}{c^3}+1}\Big)+\dfrac{a}{b^3}\Big(\frac{1}{2}-\frac{1}{\frac{b^3}{a^3}+1}\Big)\Big)$$
Now use rearrangement inequality with the constraint $\frac{c}{b}\geq \frac{a}{c}\geq \frac{b}{a}$ and $\frac{a}{b^3}\geq \frac{b}{c^3}\geq \frac{c}{a^3}$
</s>