The answer is not 0

We employ the $uvw$ method :

$\displaystyle \begin{cases} a + b + c = 3u \\ ab + bc + ca = 3v^2 \\ abc = w^3 \end{cases}$

Hence, our inequality shows that $f\left(w^3\right) \geq 0$ , where $f$ is a decreasing function.

Now, we know that $a, b, c$ are the positive roots to

$\displaystyle \begin{aligned} (x - a)(x - b)(x - c) & = 0 \\ \implies x^3 - 3ux^2 + 3v^2x - w^3 & = 0 \\ \implies x^3 - 3ux^2 + 3v^2x & = w^3 \end{aligned}$

This implies that $f(x) = x^3 - 3ux^2 + 3v^2x$ and $y = w^3$ intersect at three points: $\left(a, f(a)\right)$ , $\left(b, f(b)\right)$ , and $\left(c, f(c)\right)$ .

Assume that $a \leq b \leq c$ . Since $u^2 \geq v^2$ , we have

$f'(x) = 3x^2 - 6xu + 3v^2 = 3\left(x^2 - 2xu + v^2\right) = 3\left(x - x_1\right)\left(x - x_2\right)$

Let $x_1 \leq x_2$ . We see that $x_1 > 0$ , and that $\left(x_1, f(x_1)\right)$ and $\left(x_2, f(x_2)\right)$ are respective maximum and minimum points. Furthermore, $w^3$ increases and $u$ and $v^2$ are constants. Thus, $w^3$ is maximized when the line $y = w^3$ touches the graph of $f$ , which happens for equality of two variables.

Since our inequality is homogeneous, we can assume that $b = c = 1$ , which gives

$\displaystyle \left(a + 2\right)\left(\frac 1a + 2\right) - 9 \geq k\left(1 - \frac{2a + 1}{a^2 + 2}\right)$

or

$\displaystyle \frac{2\left(a^2 + 2\right)}{a} \geq k$

By AM-GM we have

$\displaystyle \frac{2\left(a^2 + 2\right)}{a} = 2\left(a + \frac 2a\right) \ \ \geq \ \ 2 \cdot 2\sqrt{2} = \boxed{4\sqrt{2}}$