How to correctly solve $\sqrt{1+x}+\sqrt{1-x}>1$? $$\sqrt{1+x}+\sqrt{1-x}>1, x\geq-1 \wedge x\leq1$$
$$\sqrt{1+x}>1-\sqrt{1-x}$$
$$1+x>1-2\sqrt{1-x}+1-x$$
$$0>-2\sqrt{1-x}+1-2x$$
$$2\sqrt{1-x}>1-2x$$


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*Lets say both sides of inequality are positive, then we can easily $2\sqrt{1-x}>1-2x$ square and arrange. What we get$\implies x^2<\frac{3}{4} \implies \frac{-\sqrt{3}}{2}<x<\frac{\sqrt{3}}{2}$Now, when we make set intersection with $x\geq-1 \wedge x\leq1$ we get the solution $\frac{-\sqrt{3}}{2}<x<\frac{\sqrt{3}}{2}$.

*We also have to look at the possibility that $1-2x$ can be negative: $0>1-2x\implies x>\frac{1}{2}$We determine the solution with $x\geq-1 \wedge x\leq1$ and we get the complete solution $\frac{1}{2}<x<1$. 
Now I want to get the solution for the original inequality and therefore I unite the final solutions from 1. and 2.case. I get the interval $$(-\frac{\sqrt{3}}{2}, 1]$$
This solution is wrong. The correct one is $[-1,1]$.   I would really appreciate if someone could check the complete process and point out where I made a mistake.  Also, is this solving process good and effective or I complicate things too much?

Note that the domain of $\sqrt {1+x}+\sqrt {1-x}$ is $[-1,1]$
On $(0,1]$ we have $\sqrt{1+x}>1$
On $[-1,0)$ we have $\sqrt {1-x}>1$
At $x=0$ we have $\sqrt {1+x}+\sqrt{1-x}=2>1$
Thus $\sqrt {1+x}+\sqrt {1-x}>1$ on its domain.
