Show algebraically that $x^2 + 2x + 2$ is always positive for all real values of $x$. Without using a calculator, solve the inequality $\frac{x^2 + 2x + 2}{6 - x - x^2} \le 0$. Hence, solve the inequality $\frac{x + 2\sqrt{x} + 2}{x + \sqrt{x} - 6} \ge 0$.

1. Show that the quadratic expression is always positive by completing the square.

$x^2 + 2x + 2 = (x+1)^2 - 1 + 2 = (x+1)^2 + 1$
Since $(x+1)^2 \ge 0$ for all real values of $x$, it follows that $(x+1)^2 + 1 \ge 1 > 0$.
Thus, $x^2 + 2x + 2$ is always positive.

2. Solve the rational inequality by determining the sign of the denominator.

For the inequality $\frac{x^2 + 2x + 2}{6 - x - x^2} \le 0$, since the numerator $x^2 + 2x + 2 > 0$, the inequality is satisfied if and only if the denominator is strictly negative (it cannot be zero).
$6 - x - x^2 < 0$
$x^2 + x - 6 > 0$
$(x+3)(x-2) > 0$
This gives the solution $x < -3 \text{ or } x > 2$.

3. Rewrite the third inequality to match the form of the second inequality.

The given inequality is $\frac{x + 2\sqrt{x} + 2}{x + \sqrt{x} - 6} \ge 0$.
Multiply the numerator and denominator by $-1$ to rearrange the denominator:
$\frac{(\sqrt{x})^2 + 2\sqrt{x} + 2}{-(6 - \sqrt{x} - (\sqrt{x})^2)} \ge 0$
$\frac{(\sqrt{x})^2 + 2\sqrt{x} + 2}{6 - \sqrt{x} - (\sqrt{x})^2} \le 0$

4. Solve the third inequality using the substitution $y = \sqrt{x}$ and the previous result.

Letting $y = \sqrt{x}$, the inequality becomes $\frac{y^2 + 2y + 2}{6 - y - y^2} \le 0$.
From the previous part, the solution is $y < -3 \text{ or } y > 2$.
Substituting back $y = \sqrt{x}$, we have:
$\sqrt{x} < -3 \text{ or } \sqrt{x} > 2$
Since $\sqrt{x} \ge 0$ for all real $x \ge 0$, $\sqrt{x} < -3$ yields no real solution.
For $\sqrt{x} > 2$, squaring both sides (since both are positive) gives $x > 4$.