A refinement of a famous inequality on the forum . It's related to a big problem Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$ .I have this (For one time I take the time to check it )

Let $a,b,c>0$ such that $a+b+c=1$ then we have :
  $$\sum_{cyc}\frac{a^4}{8a^3+5b^3}> \sum_{cyc}\frac{\tan(a^4)}{8\tan(a^3)+5\tan(b^3)}\geq \frac{3\tan\Big(\frac{1}{81}\Big)}{13\tan\Big(\frac{1}{27}\Big)} $$

The difficulty exceeds the level of an Olympiad I think . Furthermore I think that we cannot use Jensen's inequality (it's not homogeneous) and Cauchy-Schwarz is really too weak .Don't tell me that $\tan(x)\geq x $ is a good approximation in this case it will be a joke . 
Maybe we can prove this kind of inequality :
$$\frac{a^4}{8a^3+5b^3}+\frac{b^4}{8b^3+5a^3}\geq \frac{\tan(a^4)}{8\tan(a^3)+5\tan(b^3)}+\frac{\tan(b^4)}{8\tan(b^3)+5\tan(a^3)}$$
But even if it's works it doesn't decide the problem .
I discouraged to use power series it's really awful.
So comments and hints are welcome but don't try it alone . 
Thanks for sharing your time and knowledge.
Update :
I think it's not so hard if we remark that we have :
$$\frac{a^4}{8a^3+5b^3}>\frac{\tan(a^4)}{8\tan(a^3)+5\tan(b^3)}$$
For $a,b>0$ and $a+b<1$
Maybe someone can prove this and prove the LHS

The inequality 
$$
f(a,b)=\frac{a^4}{8a^3+5b^3}>\frac{\tan(a^4)}{8\tan(a^3)+5\tan(b^3)}
$$
reduces to
$$
5 a^4 \tan \left(b^3\right)-\left(8 a^3+5 b^3\right) \tan \left(a^4\right)+8 a^4 \tan \left(a^3\right) \ge 0
$$
in $$T=\{(a,b):a>0,b>0,a+b \le 1\}.$$
One way to do it is to cut $T$ into pieces and approximate $tan(x)$ with different upper and lower bounds.
For example, let
$$
T_1=\{(a,b):1/2 \ge a>0,b>0,a+b \le 1\}.
$$
Then we can use 
$$
\frac{x^3}{3}+x \le \tan(x) \le 2 \frac{x^3}{3}+x
$$
for $0 <x <1/2$.
For
$$
T_2=\{(a,b):1 \ge a>9/10, 1/10>b>0,a+b \le 1\},
$$
we use
$$
\frac{b^3}{3}+b \le \tan(b^3)
$$
and (Pade Approximation)
$$
\frac{-\frac{7 (a-1)^2 \left(1+\tan ^2(1)\right)}{1+3 \tan (1)}+\frac{(a-1) \left(9-9 \tan ^2(1)+17 \tan (1)\right)}{3 (1+3 \tan (1))}+\tan (1)}{\frac{(a-1) \left(-20-54 \tan ^4(1)-36 \tan ^3(1)-74 \tan ^2(1)-36 \tan (1)\right)}{6 \left(1+3 \tan ^3(1)+\tan ^2(1)+3 \tan (1)\right)}+1} \le \tan(a^3)
$$
and
$$
\frac{\frac{1}{2} (a-1) (8-3 \tan (1))+\tan (1)}{\frac{1}{2} (a-1) (-3-8 \tan (1))+1}
\ge \tan(a^4)
$$
We can continue doing this for
$$
T_3=\{(a,b):9/10 \ge a>1/2, 1/10>b>0,a+b \le 1\},
$$
and
$$
T_4=\{(a,b):9/10 \ge a>1/2, 1/10<b,a+b \le 1\}.
$$
But a slightly quicker way is to check (rigorously) that there's no critical point inside $T_3 \cup T_4$ using IntervalRootFinding.jl. Thus we just need to verify that $f(a,b)\ge 0$ on the boundary of the $T_3$ and $T_4$.
