If $abc=1$ so $\sum\limits_{cyc}\frac{7-6a}{2+a^2}\geq1$ 
Let $a$, $b$ and $c$ be real numbers such that $abc=1$. Prove that:
  $$\frac{7-6a}{2+a^2}+\frac{7-6b}{2+b^2}+\frac{7-6c}{2+c^2}\geq1$$

The equality occurs also for $a=b=2$ and $c=\frac{1}{4}$.
This inequality is a similar to the very many contest's inequalities, but nothing helps. 
At least, I don't see how we can prove it.
An example of my trying.
We need to prove that 
$$\sum_{cyc}\frac{7-6a}{2+a^2}\geq1$$ or
$$\sum_{cyc}\left(\frac{7-6a}{2+a^2}+1\right)\geq4$$ or
$$\sum_{cyc}\frac{(a-3)^2}{2+a^2}\geq4.$$
By C-S $$\sum_{cyc}\frac{(a-3)^2}{2+a^2}=\sum_{cyc}\frac{(a-3)^2(a+k)^2}{(2+a^2)(a+k)}\geq\frac{\left(\sum\limits_{cyc}(a-3)(a+k)\right)^2}{\sum\limits_{cyc}(2+a^2)(a+k)^2}$$
Now we'll  find a value of $k$, for which the equality in the last inequality occurs for $a=b=2$ and $c=\frac{1}{2}$.
Since in all equality case we have
$$\frac{a-3}{(2+a^2)(a+k)}=\frac{b-3}{(2+b^2)(b+k)}=\frac{c-3}{(2+c^2)(a+k)},$$
we obtain:
$$\frac{2-3}{(2+2^2)(2+k)}=\frac{\frac{1}{4}-3}{(2+\left(\frac{1}{4}\right)^2)(\frac{1}{4}+k)},$$
which gives $k=-\frac{9}{4}$.
Thus, it remains to prove that
$$\left(\sum\limits_{cyc}(a-3)(4a-9)\right)^2\geq4\sum_{cyc}(2+a^2)(4a-9)^2,$$
which is wrong for $a=4$ and $b=c=\frac{1}{2}$.
Any hint?
Thank you!

Clearly, we only need to prove the case $a, b, c > 0$.
Let $f(x) = \frac{7-6x}{x^2+2}.$
We have $f'(x) = \frac{(6x+4)(x-3)}{(x^2+2)^2}$.
Thus, $f(x)$ is strictly decreasing on $(0, 3)$
and strictly increasing on $(3,+\infty).$
Also, since $f(x) + 1 = \frac{(x-3)^2}{x^2+2}\ge 0$, we have $f(x)\ge -1.$
WLOG, assume that $c\le b\le a$. Clearly $c\le 1$.
If $c < \frac{1}{9}$, we have $f(a)+f(b)+f(c) \ge (-1)+(-1) + f(\frac{1}{9}) > 1.$
If $\frac{1}{9} \le c \le 1$, we have $1\le ab \le 9$. Denote $p = \sqrt{ab} \in [1, 3]$. We have
\begin{align*}
&\frac{7-6a}{a^2+2} + \frac{7-6b}{b^2+2} - \frac{2(7-6\sqrt{ab})}{ab + 2}\\
=\ &\frac{7-6a}{a^2+2} + \frac{7-6(\frac{p^2}{a})}{(\frac{p^2}{a})^2+2} - \frac{2(7-6p)}{p^2 + 2}\\
=\ & \frac{(a-p)^2[(7p^2+24p-14)a^2+(-6p^4+14p^3+24p^2-28p-24)a+7p^4+24p^3-14p^2] }{(a^2+2)(p^4+2a^2)(p^2+2)}\\
\ge\ & 0,
\end{align*}
since $7p^2+24p-14 > 0$, $7p^4+24p^3-14p^2 > 0$ and
\begin{align*}
&4(7p^2+24p-14)(7p^4+24p^3-14p^2) - (-6p^4+14p^3+24p^2-28p-24)^2\\
=\ & -12(3p^2+4p-6)(p^2-6p-2)(p^2+2)^2 > 0.
\end{align*}
Thus, it suffices to prove that $$ \frac{7-6c}{c^2+2} + \frac{2(7-6\sqrt{ab})}{ab + 2} \ge 1$$ or
$$\frac{7-6c}{c^2+2} + \frac{2(7-6\sqrt{\frac{1}{c}})}{\frac{1}{c} + 2} \ge 1$$ or
$$\frac{(3c+6\sqrt{c}+5)(2\sqrt{c}-1)^2(\sqrt{c}-1)^2}{(c^2+2)(2c+1)} \ge 0.$$
This completes the proof.
