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.
 A: 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
$$
\sin a+\cos a\le \sqrt{2}
$$
A: If @Zev Chonoles guessed right, you can prove this by computing
$$\max_{x\in [0,2\pi]} \sin x + \cos x$$
The critical points are precisely those where $\cos x - \sin x = 0$ (derivative). These are $x = \frac\pi4$ and $x = \frac{5\pi}4$. Evaluating at those and the boundary points gives
$$\sin \frac\pi4 + \cos \frac\pi4 = \frac1{\sqrt2} + \frac1{\sqrt2} = \sqrt2 \\
\sin \frac{4\pi}4 + \cos \frac{5\pi}4 = -\frac1{\sqrt2} - \frac1{\sqrt 2} = -\sqrt2 \\
\sin 0 + \cos 0 = 0 + 1 = 1\\
\sin 2\pi + \cos 2\pi = 0 + 1 = 1$$
Thus the maximum is $\max(\sqrt 2, -\sqrt 2, 1, 1) = \sqrt2$, wich proves
$$\sin x + \cos x \le \sqrt2 \qquad\forall\ x\in[0,2\pi]$$
Now since the function is $2\pi$-periodic, we can conclude
$$\sin x + \cos x \le \sqrt2 \qquad \forall\ x\in\mathbb R$$
