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!
 A: Clearly, that we only need to prove this inequality for case:$\ a , b, c >0$
Let $a=e^{t_1}, b= e^{t_2}, c= e^{t_3} $
$$f(t_1)+f(t_2)+f(t_3) \ge1 \ , \ t_1+t_2+t_3=0$$
$$f(t)=\dfrac{7-6e^t}{2+e^{2t}}$$
$f'(t)=\dfrac{2e^t(e^{t}-3)(3e^t+2)}{(e^{2t}+2)^2}$
Minimum of $f(t)$ is attained at $t=\ln(3)>0 \Rightarrow \min\left( f(t_1)+f(t_2)+f(t_3)\right)$,  for $t_1+t_2+t_3=0$ is attained only for case : $t_1,t_2,t_3 \le t_*$
Since
$f''(t)=-\dfrac{2e^t(e^t-3+\sqrt{11})(e^t-3-\sqrt{11})(3e^t+2-\sqrt{22})(3e^t+2+\sqrt{22})}{9(e^{2t}+2)^3}$
we only need to consider the inequality in case : $t_1\le t_2=t_3 \le t_*$
$a={q^2} \ , \ b=c=\dfrac{1}{q}$
$\Leftrightarrow \dfrac{(3q^2+6q+5)(2q-1)^2(q-1)^2}{(2q^2+1)(q^4+2)}\ge 0$
Equality holdes for: $(a=b=c=1)\ $  and $\ (a=\dfrac{1}{4}, b=c=2)$
A: 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.
A: There is also the following.
Let two variables be negative.
Let $a<0$, $b<0$,$a'=-a$ and $b'=-b$.
Thus, $a'>0$, $b'>0$, $a'b'c=1$ and
$$\sum_{cyc}\frac{7-6a}{2+a^2}=\frac{7+6a'}{2+a'^2}+\frac{7+6b'}{2+b'^2}+\frac{7-6c}{2+c^2}>$$
$$>\frac{7-6a'}{2+a'^2}+\frac{7-6b'}{2+b'^2}+\frac{7-6c}{2+c^2},$$ which says that it's enough to prove our inequality for positive variables.
Now, in the starting post we saw that it's enough to prove that
$$\sum_{cyc}\frac{(a-3)^2}{2+a^2}\geq4,$$ which after substitution $a=\frac{1}{x}$, $b=\frac{1}{y}$ and $c=\frac{1}{z}$ gives
$$\sum_{cyc}\frac{(3x-1)^2}{2x^2+1}\geq4.$$
Now, by C-S
$$\sum_{cyc}\frac{(3x-1)^2}{2x^2+1}\geq\frac{\left(\sum\limits_{cyc}(3x-1)\right)^2}{\sum\limits_{cyc}(2x^2+1)}.$$
Thus, it's enough to prove that:
$$\frac{\left(\sum\limits_{cyc}(3x-1)\right)^2}{\sum\limits_{cyc}(2x^2+1)}\geq4$$ or
$$9(x+y+z-1)^2\geq8(a^2+b^2+c^2)+12.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, we need to prove that
$$9(3u-w)^2\geq8(9u^2-6v^2)+12w^2,$$ which is a linear inequality of $v^2,$ which says that it's enough to prove the last inequality for an extreme value of $v^2$, which happens for equality case of two variables.
Let $y=x$ and $z=\frac{1}{x^2}.$
Id est, we need to prove that
$$9\left(2x+\frac{1}{x^2}-3\right)^2\geq8\left(2x^2+\frac{1}{x^4}\right)+12$$ or
$$(2x-1)^2(x-1)^2(x+1)(5x+1)\geq0$$ and we are done!
A: 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.
