An algebra problem by Ricky Huang

$\begin{aligned} x \left(8\sqrt{1-x} + \sqrt{1+x} \right) & \le 11 \sqrt{1+x} - 16 \sqrt{1-x} & \small \blue{\text{Divide both sides by }\sqrt{1-x}} \\ x \left(8 + \sqrt{\frac {1+x}{1-x}}\right) & \le 11 \sqrt{\frac {1+x}{1-x}} - 16 & \small \blue{\text{Let }u = \sqrt{\frac{1+x}{1-x}} \implies x = \frac {u^2-1}{u^2+1}} \\ \frac {u^2-1}{u^2+1}(8+u) & \le 11u - 16 \\ (u^2-1)(8+u) & \le (u^2+1)(11u - 16) \\ u^3 + 8u^2 - u - 8 & \le 11u^3 - 16u^2 + 11u - 16 \\ 10u^3 - 24u^2 + 12u - 8 & \ge 0 \\ 5u^3 - 12u^2 + 6u - 4 & \ge 0 & \small \blue{\text{By rational root theorem}} \\ (u-2)(5u^2-2u+2) & \ge 0 \\ \implies u & \ge 2 & \small \blue{\text{As } 5u^2 - 2u+2 = 0 \text{ has no real root.}} \\ \implies x & \ge \frac {4-1}{4+1} = 0.6 \end{aligned}$

Since $x = \dfrac {u^2-1}{u^2+1} = 1 - \dfrac 2{u^2+1}$ , implying that $x$ has a maximum value of $1$ as $u \to \infty$ , which is obvious, as $x > 1$ , the inequality will not be valid. Therefore, the inequality holds for $\boxed{0.6 \le x \le 1}$ .

Since $x$ is real, it can't be greater than $1$ . So $x\leq 1$

Simplifying the given equation we get the cubic equation

$65x^3+171x^2+99x-135\geq 0$

This inequality has only one real solution $x\geq 0.6$

Therefore the correct option is

$\boxed {0.6\leq x\leq 1}$ .