Prove inequality with complicated square roots I've been trying to prove the following inequality but the square roots really give me a hard time.
$$ x \cdot (\sqrt{4-x^2} +2x -\sqrt2)\leq\sqrt{-(x^2-4x+2)(x^2+4x+2)} $$ for $ 1 \leq x\leq\sqrt2$. Considering both sides as a function and using WolframAlpha,I found the intersection points $ x= \frac{\sqrt2}{2} $ and $ x=\sqrt2 $ but I couldn't prove that there is no other intersection point within this interval. Since the derivatives are also quite complicated I'm stuck.

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.
