Force Out The Inequality!

Since the above expression is negative when $x<0$ and positive when $x>0$ , the maximum value will occur when $x$ is positive. So assume $x>0$ .

Note that $y = \frac{x^2+2-\sqrt{x^4+4}}{x}=x+\frac{2}{x}-\sqrt{x^2+\frac{4}{x^2}}$

Let $t=x^2+\frac{4}{x^2}$ . By $\text{AM-GM}$ we have that $t \ge 4$ .

Then $y=\sqrt{t+4}-\sqrt{t}=\frac{4}{\sqrt{t+4}+\sqrt{t}}$

So we have to minimize $\sqrt{t+4}+\sqrt{t}$ and since $t \ge 4$ the minimum will be $2\sqrt{2}+2$ .

So the minimum of $y$ is $\frac{4}{2\sqrt{2}+2}=2\sqrt{2}-2$ or approximately $0.828$ .

Let $y=\dfrac{x^2+2-\sqrt{x^4+4}}x$

Note that $x\neq 0$

Cross - Multiplying and Rearranging, $\sqrt{x^4+4} = x^2-xy+2\\\Rightarrow x^4+4 = (x^2-xy+2)^2\\\Rightarrow 2x^3y-x^2(y^2+4)+4xy=0\\\Rightarrow x(2yx^2-(y^2+4)+4y)=0$

Since $x\neq 0$ , the above solution should have real roots for $x$ . So, $\Delta = (y^2+4)^2-4\cdot4y\cdot2y= (y^2+4)^2-32y^2\ge0$

So, $y^2\le4(3-2\sqrt2) ~\text{or}~4(3+2\sqrt2)\le y^2$

Since we take negative value of $\sqrt{x^2+4}$ , we take the that condition which gives maximum value. So, $y^2\le4(3-2\sqrt2)\\|y|<2(\sqrt2-1)$

Hence maximum value of $y$ is $\color{#3D99F6}{2(\sqrt2-1)\approx \boxed{0.828427}}$