Show $\frac1{a^2+a+1}+\frac1{b^2+b+1}+\frac1{c^2+c+1}\ge1$ Positive real numbers $a,b,c$ satisfy $abc=1$. Prove
  $$\frac1{a^2+a+1}+\frac1{b^2+b+1}+\frac1{c^2+c+1}\ge1$$


I tried AM&ndash;GM, Cauchy&ndash;Schwarz and Jensen's but they all failed.
Following stewbasic's hint in the comments:
$$\frac1{a^2+a+1}+\frac1{b^2+b+1}+\frac1{c^2+c+1}\ge1$$
$$\iff(b^2+b+1)(c^2+c+1)+(c^2+c+1)(a^2+a+1)+(a^2+a+1)(b^2+b+1)\ge(a^2+a+1)(b^2+b+1)(c^2+c+1)$$
The LHS expands as
$$a^2b^2+a^2b+ab^2+\color{blue}{a^2}+b^2+ab+a+b+1\\
+b^2c^2+b^2c+bc^2+\color{blue}{b^2}+c^2+bc+b+c+1\\
+c^2a^2+c^2a+ca^2+\color{blue}{c^2}+a^2+ca+c+a+1$$
while the RHS expands as
$$a^2b^2c^2+abc(ab+bc+ca)+\\
a^2b^2+b^2c^2+c^2a^2+abc(a+b+c)+\\
abc+a^2b+b^2a+b^2c+c^2b+c^2a+a^2c+\\
\color{green}{ab+bc+ca}+a^2+b^2+c^2+a+b+c+1$$
Cancelling and using the relation $abc=1$ we find that the inequality above is equivalent to
$$\color{blue}{a^2+b^2+c^2}\ge\color{green}{ab+bc+ca}$$
which can be seen to be true by applying the Cauchy&ndash;Schwarz inequality on $(a,b,c)$ and $(b,c,a)$. By following the biconditionals back we see that the original inequality is true.
Let $a=\frac{x}{y}$, $b=\frac{y}{z}$ and $c=\frac{z}{x}$, where $x$, $y$ and $z$ are positives.

Hence, by C-S and Rearrangement we obtain $$\sum\limits_{cyc}\frac{1}{a^2+a+1}=\sum\limits_{cyc}\frac{y^2}{x^2+xy+y^2}=\sum\limits_{cyc}\frac{y^2(y+z)^2}{(x^2+xy+y^2)(y+z)^2}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}(x^2+xy)\right)^2}{\sum\limits_{cyc}(x^2+xy+y^2)(y^2+2yz+z^2)}=\frac{\sum\limits_{cyc}(x^4+2x^3y+2x^3z+3x^2y^2+4x^2yz)}{\sum\limits_{cyc}(x^4+2x^3y+x^3z+3x^2y^2+5x^2yz)}=$$
$$=\frac{\sum\limits_{cyc}(x^4+2x^3y+x^3z+3x^2y^2+5x^2yz)+xyz\left(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}-x-y-z\right)}{\sum\limits_{cyc}(x^4+2x^3y+x^3z+3x^2y^2+5x^2yz)}\geq1$$