<s>[INST] 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!
 [/INST] 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!
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