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 \(\sqrt{1+x}+\sqrt{1-x}<2\).