For positive $a$, $b$, $c$ with $abc=1$, show that $\sum_{cyc}\sqrt{a^2-a+1}\geq a+b+c$

The hint:

Use the Mixing Variables method.

Indeed, we can use the beautiful Can's idea.

Since $$\prod\limits_{cyc}(a-1)^2=\prod_{cyc}((a-1)(b-1))\geq0,$$ we can assume that $$(a-1)(b-1)\geq0$$ or $$a+b\leq1+ab=1+\frac{1}{c}.$$

Thus, by C-S:

\begin{align} \sqrt{a^2-a+1}&+\sqrt{b^2-b+1}\\ &=\sqrt{a^2+b^2-a-b+2+2\sqrt{(a^2-a+1)(b^2-b+1)}}\\ &\geq\sqrt{a^2+b^2-a-b+2+2\sqrt{\left(\left(a-\frac{1}{2}\right)^2+\frac{3}{4}\right)\left(\left(b-\frac{1}{2}\right)^2+\frac{3}{4}\right)}}\\ &\geq\sqrt{a^2+b^2-a-b+2+2\left(\left(a-\frac{1}{2}\right)\left(b-\frac{1}{2}\right)+\frac{3}{4}\right)}\\ &=\sqrt{a^2+b^2-a-b+2+2ab-a-b+2}\\ &=\sqrt{(a+b)^2-2(a+b)+4}. \end{align}

But $f(x)=\sqrt{x^2-2x+4}-x$ decreases, which says $$\sum_{cyc}(\sqrt{a^2-a+1}-a)\geq f(a+b)+\sqrt{c^2-c+1}-c\geq$$ $$\geq f\left(1+\frac{1}{c}\right)+\sqrt{c^2-c+1}-c=\sqrt{3+\frac{1}{c^2}}-1-\frac{1}{c}+\sqrt{c^2-c+1}-c.$$ Id est, it's enough to prove that: $$\sqrt{3+\frac{1}{c^2}}-1-\frac{1}{c}+\sqrt{c^2-c+1}-c\geq0$$ and the rest is smooth.

Can you end it now?