<s>[INST] Prove that $\sin (\theta) + \cos(\theta) \ge 1$ 
Let $\theta$ be an arbitrary acute angle. Prove that  $\sin (\theta) + \cos(\theta) \ge 1$.

$$\big(\sin (\theta) + \cos (\theta)\big)^2 = 1 + 2 \sin(\theta)\cos(\theta)\ge 0$$
so, \begin{align*}\big(\sin(\theta)) + \cos(\theta)\big)^2 &> 1\\ 
\big(\sin(\theta)+ \cos(\theta)\big)^2 &\ge 1\end{align*}
 [/INST] This is a brute force approach.
There is an $R$ and $\alpha$ such that $\sin \theta + \cos \theta = R\sin(\theta + \alpha)$.
First, $R\sin(\theta + \alpha) = R\cos\alpha\sin\theta + R\sin\alpha\cos\theta$.
So $R\sin\alpha = R\cos\alpha = 1$. Since $\tan\alpha = 1$, $\alpha = 45^\circ$. Adding the squares of $R\sin\alpha$ and $R\cos\alpha$ gets $R^2 = 2 \therefore R = \sqrt 2$.
$$\sin\theta + \cos\theta = \sqrt{2}\sin(\theta + 45^\circ)$$
It should be clear now if you look at a graph of the $\sin$ function.

Using complex numbers:
$$e^{i\theta} - ie^{i\theta} = (\cos\theta + \sin\theta) + i(\sin\theta - \cos\theta) = (1-i)e^{i\theta} = \sqrt{2}e^{i(\theta - 45^\circ)}$$
Now trace the $90^\circ$ arc from $\theta = 0$ to $\theta = 90^\circ$, and it's clear that the real part is bigger than $1$.
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