You can use QM-AM for $\sqrt{a},\ \sqrt{b},\ \sqrt{c},\ \sqrt{d}$; I'm leaving the details for you.
Let
$$A := \frac{x^2 + 2}{4x}$$
and
$$B := \frac{-x}{2(\sqrt{-2(x^2-4)}-3)} = \frac{x}{6 - 2\sqrt{8 - 2x^2}} = \frac{x(6 + 2\sqrt{8 - 2x^2})}{8x^2 + 4}.$$
We have
$$\sqrt{8 - 2x^2} = \frac{(8-2x^2) - 2^2}{\sqrt{8-2x^2} + 2} + 2 \le \frac{4 - 2x^2}{2 + 2} + 2 = 3 - x^2/2$$
and
$$B \le \frac{x[6 + 2(3 - x^2/2)]}{8x^2 + 4} =
\frac{x(12 - x^2)}{8x^2 + 4} =: C.$$
It suffices to prove that
$$x \le \sqrt{1 - A^2} + \sqrt{1 - C^2}$$
or
$$x/2 - \sqrt{1 - C^2} \le \sqrt{1 - A^2} - x/2$$
or
$$\frac{(x/2)^2 - (1 - C^2)}{x/2 + \sqrt{1 - C^2}} \le \frac{1 - A^2 - (x/2)^2}{\sqrt{1 - A^2} + x/2}.$$
We have
$$1 - A^2 - (x/2)^2 = \frac{(2 - x^2)(5x^2 - 2)}{16x^2} \ge 0 \tag{1}$$
and
$$(x/2)^2 - (1 - C^2) = \frac{(2-x^2)^2(17x^2-4)}{16(2x^2+1)^2} \ge 0. \tag{2}$$
It is easy to prove that $C \le \frac{11}{12}$.
Thus, we have
$$\sqrt{1 - C^2} \ge \sqrt{1 - \frac{121}{144}}
> \frac{\sqrt 2}{5} \ge \frac{x}{5}. \tag{3}$$
Using AM-GM, we have $A \ge \frac{2\sqrt{x^2 \cdot 2}}{4x} = \frac{\sqrt 2}{2}$. Thus, we have
$$\sqrt{1 - A^2} \le \sqrt{1 - 1/2} < 3/4 \le 3x/4. \tag{4}$$
Using (1)-(4), it suffices to prove that
$$\frac{1}{x/2 + x/5}\cdot \frac{(2-x^2)^2(17x^2-4)}{16(2x^2+1)^2} \le \frac{1}{3x/4 + x/2}\cdot \frac{(2 - x^2)(5x^2 - 2)}{16x^2}.$$
Using $2x^2 + 1 - 3x = (2x - 1)(x - 1) \ge 0$, it suffices to prove that
$$\frac{1}{x/2 + x/5}\cdot \frac{(2-x^2)^2(17x^2-4)}{16\cdot (3x)^2} \le \frac{1}{3x/4 + x/2}\cdot \frac{(2 - x^2)(5x^2 - 2)}{16x^2}$$
or
$$25(2 - x^2)(17x^2 - 4) \le 126(5x^2 - 2)$$
which is true. (Hint: Let $y = x^2 \in [1, 2]$.)
We are done.
Best Answer
Let $P := -(x^2 - 4x + 2)(x^2 + 4x + 2) = (4x)^2 - (x^2 + 2)^2$.
It suffices to prove that $$x(\sqrt{4 - x^2} - \sqrt 2) \le \sqrt{P} - 2x^2$$ or $$x\cdot \frac{2 - x^2}{\sqrt{4 - x^2} + \sqrt 2 } \le \frac{P - (2x^2)^2}{\sqrt{P} + 2x^2}$$ or $$x\cdot \frac{2 - x^2}{\sqrt{4 - x^2} + \sqrt 2 } \le \frac{(2 - x^2)(5x^2 - 2)}{\sqrt{P} + 2x^2}.$$
Since $\sqrt{4 - x^2} + \sqrt 2 \ge \sqrt{4 - 2} + \sqrt 2 > 2$ and $$P \le (4x)^2 - 4 \cdot x^2 \cdot 2 = 8x^2 \le 9x^2,$$ it suffices to prove that $$x\cdot \frac{2 - x^2}{2} \le \frac{(2 - x^2)(5x^2 - 2)}{3x + 2x^2}$$ or $$x/2 \le \frac{5x^2 - 2}{3x + 2x^2}$$ or $$-2x^3+7x^2-4 \ge 0$$ or $$(x-1)(-2x^2+5x+5) + 1 \ge 0$$ which is true.
We are done.