Can you square both sides in a proof? Like say we had to prove that $\sqrt{2} + \sqrt{3} < \sqrt{26}$. Could you square both sides to prove it?
 A: Yes, you can, since both numbers are non-negative. That is, if $a,b\in[0,\infty)$, then $a<b\iff a^2<b^2$.
A: We have: $\sqrt{2} + \sqrt{3} < 2\sqrt{3} = \sqrt{12} < \sqrt{26}$. That's probably faster.
A: Expanding on José Carlos Santos's answer:  yes, in this case, but not generally.  He shows the "yes, in this case" part.  Generally, however, signs matter.  From the true statement $-1 < 1$, squaring both sides seems to give $1 \overset{?}{<} 1$, which is not true.
A: In general, if $f(x)$ is any strictly increasing function over any interval $I = (a,b)$ (or $[a,b), (a,\infty), (-\infty,b),...$), then for any $u,v \in I$, $u < v \iff f(u) < f(v)$.
Taking square $x \mapsto x^2$ and square roots $x \mapsto \sqrt{x}$ are strictly increasing functions over $[0,\infty)$. So for any non-negative $u, v$, you have 
$$u < v \iff u^2 < v^2 \iff \sqrt{u} < \sqrt{v}$$ 
For your case, you know both sides are non-negative.
This means
$$\begin{align} & \sqrt{2} + \sqrt{3} \stackrel{?}{<} \sqrt{26}\\
\iff & 5 + 2\sqrt{6} = (\sqrt{2}+\sqrt{3})^2 \stackrel{?}{<} 26\\
\iff & 2\sqrt{6} \stackrel{?}{<} 21\\
\iff & 24 = (2\sqrt{6})^2 < 21^2 = 441
\end{align}
$$
Since you know the last inequality is true, the inequality you want to show $\sqrt{2} + \sqrt{3} < \sqrt{26}$ is also true.
