Review question

Can we prove this inequality involving two square roots? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7533

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