<s>[INST] Prove the inequality $b\le \sqrt{a^2 + b^2}$ 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.
 [/INST] It is not true that $b =\sqrt {b^{2}}$. You have to say $b \leq |b| =\sqrt {b^{2}} \leq \sqrt {a^{2}+b^{2}}$. 
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