Given three positive numbers $a,b,c\in R_{+}^{*}$ prove the following inequality 
If  $\ abc=1$  then prove that
  $$\sum_{cyc}\frac{1}{3-a+a^{6}}≤1$$
  where $a,b,c>0$

I think this inequality can be proved by holder ? 
My attempt using  $\ am-gm$
$$3-a+a^{6}≥3-a \quad(etc)$$
$$\sum_{cyc}\frac{1}{3-a+a^{6}}≤\displaystyle\sum_{cyc}\frac{1}{3-a}$$
Now I will get tow case if $a>3$ and if $a<3$.
If $a>3,$ the  inequality is true.
Now if $a<3$ 
Using :  $3-a>0$ (etc)
So: $$\sum_{cyc}\frac{1}{3-a}≤1$$
Is my work correct? 
 A: $$\sum_{cyc}\frac{1}{a^6-a+3}\leq1$$ it's
$$\sum_{cyc}\left(\frac{1}{3}-\frac{1}{a^6-a+3}\right)\geq0$$ or
$$\sum_{cyc}\left(\frac{1}{3}-\frac{1}{a^6-a+3}-\frac{5}{9}\ln{a}\right)\geq0.$$
Let $f(x)=\frac{1}{3}-\frac{1}{x^6-x+3}-\frac{5}{9}\ln{x}.$
Thus, $$f'(x)=\tfrac{(1-x)(5x^{11}+5x^{10}+5x^9+5x^8+5x^7-5x^6-29x^5-29x^4-29x^3-29x^2-24x-45)}{9x(x^6-x+3)^2}.$$
We see that the polynomial $x^{11}+5x^{10}+5x^9+5x^8+5x^7-5x^6-29x^5-29x^4-29x^3-29x^2-24x-45$
has only one changing of coefficients sign, which by the Descartes's rule https://en.wikipedia.org/wiki/Descartes%27_rule_of_signs
says that this polynomial has unique positive root $x_1$ and easy to see that $x_1>1$.
For $x=1$ our $f$ has a local minimum and for $x=x_1$ our $f$ has a local maximum,
$f(x_1)>0$ and $f$ decreases on $[x_1,+\infty).$
Since $\lim\limits_{x\rightarrow+\infty}f(x)=-\infty,$ we obtain that $f$ has an unique root $x_0$ on  $[x_1,+\infty)$.
By calculator easy to see that $x_0=1.696...$ and since $f(1)=0$, 
we got that our inequality is proven for $\max\{a,b,c\}\leq1.696.$ 
Now, let $a>1.696.$
Thus, by AM-GM $$\sum_{cyc}\frac{1}{a^6-a+3}<\frac{1}{1.696^6-1.696+3}+\frac{2}{3-\frac{5}{6\sqrt[5]6}}=0.867...<1$$ and we are done!
