How to correctly solve $\sqrt{1+x}+\sqrt{1-x}>1$?

Question

$$\sqrt{1+x}+\sqrt{1-x}>1, x\geq-1 \wedge x\leq1$$ $$\sqrt{1+x}>1-\sqrt{1-x}$$ $$1+x>1-2\sqrt{1-x}+1-x$$ $$0>-2\sqrt{1-x}+1-2x$$ $$2\sqrt{1-x}>1-2x$$

1. 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 $\frac{-\sqrt{3}}{2}<x<\frac{\sqrt{3}}{2}$.

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?

Answer 1

Note that if $x > 0$, then $\sqrt{1 + x} > 1$; similarly, when $x \leq 0$, we have that $\sqrt{1 - x} > 1$. As we have a sum of two positive numbers, of which at least one is larger or equal to one, on the LHS, it must always be larger than one.

Because of the existence of this particular argument, your solving process is too complicated indeed. However, removing square roots from an equation by squaring isn't a bad tactic in se. Just make sure the inequalities still hold after squaring.

Answer 2

The domain of this inequation is $[-1,1]$. On its domain, both sides are non-negative, so comparing them amounts to comparing their squares: $$\bigl(\sqrt{1+x}+\sqrt{1-x}\bigr)^2=1+\not x+1-\not x+2\sqrt{1-x^2}>1\iff 1+2\sqrt{1-x^2}>0,$$ which is satisfied by any $x$ in the domain since $1+2\sqrt{1-x^2}\ge 1$.

Answer 3

Hint.-$f(x)=\sqrt{1+x}+\sqrt{1-x}$ and $f(x)=f(-x)$ and $f'(x)\lt0$ when $x\gt0$. Then the minimum of $f(x)$ is $f(1)=\sqrt2\gt1$.

Thus $f(x)\gt1$ on all its domain $[-1,1]$

Answer 4

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.