Because for $x > 0$ we have that $\frac{2}{x^2}-\frac{1}{x}-1 \ge 2-x-x^2$ Sum up cyclic.

Positive real numbers $a, b, c$ satisfy $a^2 + b^2 + c^2 + a + b + c = 6$. Prove the following inequality: $$k(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}) \geq 3(k-1) + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$$Where $ k\geq 1 .$