A big problem we get around $(x,y,z)=(0.822,1.265,1.855)$.
The Buffalo Way helps:
Let $x=\min\{x,y,z\}$, $y=x+u$,$z=x+v$ and $x=t\sqrt{uv}$.
Hence, $\frac{13}{5}\prod\limits_{cyc}(8x^3+5y^3)\left(\sum\limits_{cyc}\frac{x^4}{8x^3+5y^3}-\frac{x+y+z}{13}\right)=$
$$=156(u^2-uv+v^2)x^8+6(65u^3+189u^2v-176uv^2+65v^3)x^7+$$
$$+2(377u^4+1206u^3v+585u^2v^2-1349uv^3+377v^4)x^6+$$
$$+3(247u^5+999u^4v+1168u^3v^2-472u^2v^3-726uv^4+247)x^5+$$
$$+3(117u^6+696u^5v+1479u^4v^2+182u^3v^3-686u^2v^4-163uv^5+117v^6)x^4+$$
$$+(65u^7+768u^6v+2808u^5v^2+2079u^4v^3-1286u^3v^4-585u^2v^5+181uv^6+65v^7)x^3+$$$$+3uv(40u^6+296u^5v+472u^4v^2-225u^2v^4+55uv^5+25v^6)x^2+ $$
$$+u^2v^2(120u^5+376u^4v+240u^3v^2-240u^2v^3-25uv^4+75v^5)x+$$
$$+5u^3v^3(8u^4+8u^3v-8uv^3+5v^4)\geq$$
$$\geq u^5v^5(156t^8+531t^7+2t^6-632t^5-152t^4+867t^3+834t^2+299t+40)\geq0$$
Done!
For example, we'll prove that $$6(65u^3+189u^2v-176uv^2+65v^3)\geq531\sqrt{u^3v^3},$$ which gives a coefficient $531$ before $t^7$ in the polynomial $156t^8+531t^7+2t^6-632t^5-152t^4+867t^3+834t^2+299t+40.$
Indeed, let $u=k^2v$, where $k>0$.
Thus, we need to prove that:
$$130k^6+378k^4-177k^3-352k^2+130\geq0$$
and by AM-GM we obtain: $$130k^6+378k^4-177k^3-352k^2+130=$$
$$=130\left(k^3+\frac{10}{13}k-1\right)^2+\frac{k}{13}(2314k^3+1079k^2-5576k+2600)\geq$$
$$\geq\frac{k}{13}\left(8\cdot\frac{1157}{4}k^3+5\cdot\frac{1079}{5}k^2+21\cdot\frac{2600}{21}-5576k\right)\geq$$
$$\geq\frac{k^2}{13}\left(34\sqrt[34]{\left(\frac{1157}{4}\right)^8\left(\frac{1079}{5}\right)^5\left(\frac{2600}{21}\right)^{21}}-5576\right)>0.$$
We'll prove that $$
2(377u^4+1206u^3v+585u^2v^2-1349uv^3+377v^4)\geq2u^2v^2,$$ for which it's enough to prove that:
$$377t^4+1206t^3+584t^2-1349t+377\geq0$$ or
$$t^4+\frac{1206}{377}t^3+\frac{584}{377}t^2-\frac{1349}{377}t+1\geq0$$ or
$$\left(t^2+\frac{603}{377}t-\frac{28}{29}\right)^2+\frac{131015t^2-69589t+9633}{142129}\geq0,$$ which is true because
$$69589^2-4\cdot131015\cdot9633<0.$$
Only a partial answer.
Assuming $a\le b\le c$, then we have $0<a\le1$ and $1\le c<3$. Now we have two cases, $b\le1$ and $b>1$. The case $b\le1$ is easy to deal with.
Assuming $0<a\le b\le1\le c<3$, we have
\begin{align}
\left(\frac{a+1}{a+b} \right)\left(\frac{b+1}{b+c}\right)\left(\frac{c+1}{c+a} \right) \ge 1
&\iff (a+1)(b+1)(4-a-b)\ge(a+b)(3-a)(3-b)\\
&\iff {\left(2-a-b\right)} {\left(1-a\right)} {\left(1-b\right)}\ge0.
\end{align}
Now assuming $0<a\le1\le b \le c<3$. Below is not an answer but only an analysis.
We want to know why this case is difficult, and why $\frac25$ is important.
Using the series expansion of $\exp$, and let $A = \operatorname{diag}\left(\ln\left(\frac{a+1}{a+b} \right),\ln\left(\frac{b+1}{b+c}\right),\ln\left(\frac{c+1}{c+a} \right)\right)$, then we have
\begin{align}
LHS
&= \sum_{n=0}^\infty\left(\frac25\right)^n\frac{\operatorname{tr}A^n}{n!}\\
&= 3 + \frac25\operatorname{tr}A + \frac12\cdot \left(\frac25\right)^2\operatorname{tr}A^2+ \frac16\cdot \left(\frac25\right)^3\operatorname{tr}A^3 +R_3,
\end{align}
where
$R_3 = \sum_{n=4}^\infty\left(\frac25\right)^n\frac{\operatorname{tr}A^n}{n!} \ge 0$, since $e^x-(1+x+x^2/2+x^3/6)$ is positive for any $x\in\mathbb R$.
Then we can prove the inequality if we have
$$\frac25\operatorname{tr}A + \frac12\cdot \left(\frac25\right)^2\operatorname{tr}A^2+ \frac16\cdot \left(\frac25\right)^3\operatorname{tr}A^3\ge\!\!\!?\;0,$$
which can be simplified to
$$75\operatorname{tr}A + 15\operatorname{tr}A^2+ 2\operatorname{tr}A^3\ge\!\!\!?\;0.\tag{1}$$
Numerical results suggest that using the 3 first terms is enough to prove the inequality. Note that in the first case where $b\le1$, using the first term $\operatorname{tr}A$ is enough (and what we did in the first part is in fact proving $\operatorname{tr}A\ge0$), that's why that case is easy.
So,
Why the case $b\ge1$ is difficult?
Because we have 2 more terms, $\operatorname{tr}A^2$ and $\operatorname{tr}A^3$, to deal with.
Why $\frac25$ is important?
Because $\frac25$ gives the coefficients 75, 15, and 2, which makes $75\operatorname{tr}A + 15\operatorname{tr}A^2+ 2\operatorname{tr}A^3\ge0$.
Best Answer
Let $x_1=a^3$, $x_2=b^3$, $x_3=c^3$, and $S=\sum_i x_i$. We wish to prove
$$ \frac{x_1}{(S-x_3)^{3/4}}+\frac{x_2}{{(S-x_1)}^{3/4}}+\frac{x_3}{{(S-x_2)}^{3/4}}-\frac{\sum_i x_i^{1/4}}{2^{3/4}}\ge 0.$$
We consider the problem to minimize $ \frac{x_1}{(S-x_3)^{3/4}}+\frac{x_2}{{(S-x_1)}^{3/4}}+\frac{x_3}{{(S-x_2)}^{3/4}}-\frac{\sum_i x_i^{1/4}}{2^{3/4}}$ over variables $x_1,x_2, x_3$, such that $x_i\ge 0$ and $\sum_i x_i = S$. We see that all $x_i$ equal satisfies KKT and is indeed the minimal choice (If Largrange multiplier for any of $x_i>0$ is non-zero, $x_i=0$ and the condition holds and thus we can assume that the Lagrange for each $x_i$ is zero and thus there is a single Lagrange multiplier for the sum and equal $x_i$ trivially satisfies the conditions.). Thus, we get the above expression to hold.