Prove the inequality $b\le \sqrt{a^2 + b^2}$

Question

How do i prove $b\le \sqrt{a^2 + b^2}$ ? What i think is :

$\begin{align} a^2& \gt 0,\, b^2 \gt 0 \\ b &= \sqrt{b^2}\\ &\le\sqrt{a^2+b^2} \\ \therefore b &\le \sqrt{a^2 + b^2} \end{align}$

Am i right? Please do a correction.

Answer 1

It is not true that $b =\sqrt {b^{2}}$. You have to say $b \leq |b| =\sqrt {b^{2}} \leq \sqrt {a^{2}+b^{2}}$.