Prove that $\sin(a)$ + $\cos(a)\leq\sqrt{2}$ $$\begin{align*}
\sin (a) + \cos(a) &\leq \sqrt{2}\\
(\sin(a)+ \cos(a))^2 &\leq (\sqrt{2})^2\\
\sin^2(a) + 2\sin(a)\cos(a) + \cos^2(a) &\leq \text{2}
\end{align*}$$
Am I doing it right? I need help.

Alternative path:
$$
\sin a+\cos a=
\sqrt{2}\left(\sin a\cos\frac{\pi}{4}+\cos a\sin\frac{\pi}{4}\right)=
\sqrt{2}\sin\left(a+\frac{\pi}{4}\right)\le\sqrt{2}
$$
(and also $\ge-\sqrt{2}$, of course).
However your reasoning is basically correct; only you need to do it backwards:
$$
\sin^2a+2\sin a\cos a+\cos^2a\le2
$$
because $\sin^2a+\cos^2a=1$ and $2\sin a\cos a=\sin 2a\le 1$; therefore
$$
(\sin a+\cos a)^2\le 2
$$
and so
$$
\sin a+\cos a\le \sqrt{2}
$$
