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!

Michael Rozenberg's solution without UVW:
Michael Rozenberg gave a nice application of Cauchy-Bunyakovsky-Schwarz inequality.
Then it suffices to prove that, for all $a, b, c > 0$ with $a b c = 1$,
$$9(a + b + c - 1)^2 - 8(a^2 + b^2 + c^2) - 12 \ge 0.$$
Denote LHS by $f(a, b, c)$. WLOG, assume $c\ge 1$. We have
$$f(a, b, c) - f(\sqrt{ab}, \sqrt{ab}, c)
= (\sqrt{a} - \sqrt{b})^2(a + 2\sqrt{ab} + b + 18c - 18) \ge 0.$$
Also, we have
$$f(\sqrt{ab}, \sqrt{ab}, c)
= \frac{(\sqrt{c} + 5)(\sqrt{c} + 1)(\sqrt{c} - 2)^2(\sqrt{c} - 1)^2}{c} \ge 0.$$
We are done.
