Prove that $\sum_{\mathrm{cyc}} \frac{214x^4}{133x^3 + 81y^3} \ge x + y + z$ for $x, y, z > 0$ Problem. Let $x, y, z > 0$. Prove that 
$$\frac{214x^4}{133x^3 + 81y^3} + \frac{214y^4}{133y^3 + 81z^3} + \frac{214z^4}{133z^3 + 81x^3} \ge x+y+z.$$
It is verified by Mathematica. The inequality holds with equality if $x = y = z$.
When $x = \frac{121}{84}, y = \frac{43}{66}$ and $z = 1$, 
$\mathrm{LHS} - \mathrm{RHS} \approx 0.000005327884220$.
It is a stronger version of the inequality in this link: 
Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$
They can be written as (for $k = \frac{8}{5}$ and $k = \frac{133}{81} \approx 1.641975$, respectively)
$$\frac{x^4}{kx^3+y^3} + \frac{y^4}{ky^3 + z^3} + \frac{z^4}{kz^3 + x^3} \ge \frac{x+y+z}{k+1}.$$
The best constant $k$ is approximately $1.64199$ (see the comment by @Colescu in the link above).
I can prove the inequality of $k = \frac{8}{5}$ by the Buffalo Way. 
Several months ago, I tried to prove the inequality of $k = \frac{133}{81}$ by the Buffalo Way without success.
However, I think that the Buffalo Way may work but just I have not found the way.
Any comments and solutions are welcome.

We can reduce a degree of this inequality.
Indeed, by C-S we obtain:
$$\sum_{cyc}\frac{x^4}{133x^3+81y^3}=\sum_{cyc}\frac{x^4(200x-57y+154z)^2}{(133x^3+81y^3)(200x-57y+154z)^2}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}(200x^3-57x^2y+154x^2z)\right)^2}{\sum\limits_{cyc}(133x^3+81y^3)(200x-57y+154z)^2}$$ and it's enough to prove that:
$$214\left(\sum\limits_{cyc}(200x^3-57x^2y+154x^2z)\right)^2\geq(x+y+z)\sum\limits_{cyc}(133x^3+81y^3)(200x-57y+154z)^2,$$ which is true, but BW does not help here and I have no a proof by hand. 
