First, the problem: let $a,\ b,\ c > 0$ be such that $abc = 1$. Prove that $$\sum_{cyc} (a^2 - 2bc + 1) \ge 0$$
Some background: I was solving a competition inequality by playing around with possible identities/other inequalities that could lead to a solution. Essentially, I was looking for possible bounds that might help and then I was checking them numerically. So, I have verified the validity of the above inequality numerically. However, I have no idea how to actually prove it. Any suggestions are welcomed.
Let $a^2=x^3$, $b^2=y^3$ and $c^2=z^3$.
Hence, $xyz=1$ and by Schur and AM-GM we obtain: $$\sum_{cyc}(a^2-2bc+1)=\sum_{cyc}(x^3-2\sqrt{x^3y^3}+xyz)=$$ $$=\sum_{cyc}(x^3-x^2y-x^2z+xyz)+\sum_{cyc}(x^2y+xy^2-2\sqrt{x^3y^3})\geq0.$$