Can we prove this inequality involving two square roots?

Can we prove this inequality involving two square roots?

Question

Given that $0<x<1$, prove that $\sqrt{1+x}+\sqrt{1-x}<2.$

Suggestion

Given that $0<x<1$, prove that $\sqrt{1+x}+\sqrt{1-x}<2$.

Can we get rid of those square roots? Would squaring both sides be helpful? Are we allowed to do this?

Solution

Given that $0<x<1$, prove that $\sqrt{1+x}+\sqrt{1-x}<2$.

We need to be careful when squaring inequalities. For example, $-3 < 2$, but $(-3)^2 > 2^2.$

But if $0 < a < b$, then we can definitely say that $0 < a^2 < b^2$ and $0 < \sqrt{a} < \sqrt{b}$.

We will use this idea to help us answer this question.

It’s true here that $0 < \sqrt{1+x}+\sqrt{1-x}$.

So $0 < \sqrt{1+x}+\sqrt{1-x}<2$ if and only if $0 < \bigl(\sqrt{1+x}+\sqrt{1-x}\bigr)^2<4$.

Therefore, to prove the inequality we require, we only need to show that $0 < \bigl(\sqrt{1+x}+\sqrt{1-x}\bigr)^2<4$.

The left hand inequality is clear.

Expanding the central expression, we have

$$\begin{align*} \bigl(\sqrt{1+x}+\sqrt{1-x}\bigr)^2&=(1+x) + 2\sqrt{(1+x)(1-x)} + (1-x)\\ &=2+2\sqrt{1-x^2}. \end{align*}$$

Since $\sqrt{1-x^2}<1$, the right hand side is less than $4$, as we wanted.

So we have $\sqrt{1+x}+\sqrt{1-x}<2$.