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*}
 A: Both $\sin(\theta)$ and $\cos(\theta)$ are concave functions on $I=\left[0,\frac{\pi}{2}\right]$, so their sum is a concave function too, and attains its minimum in the endpoints of $I$.
$$\forall \theta\in I,\quad f(\theta)=\sin\theta+\cos\theta \geq f(0) = f\left(\frac{\pi}{2}\right) = 1 $$
follows.
A: You already had it almost and in a very elementary way: since we're given that $\;0<\theta<\frac\pi2\;$ , we have that
$$\begin{cases}\sin\theta >0\\{}\\\cos\theta>0\end{cases}\;\;\implies \;\sin\theta+\cos\theta\stackrel{\color{red}{(**)}}>0\;,\;\;\text{and thus:}$$
$$\left(\sin\theta+\cos\theta\right)^2=1+2\sin\theta\cos\theta>1\stackrel{\color{red}{(**)}}\implies \sin\theta+\cos\theta>1$$
If you want weak inequality in the first line above then you get weak inequality in the last lines, too.
A: For a geometric approach: Draw a right triangle with hypotenuse $1$.  Mark one of the acute angles as $\theta$.  Then $\sin(\theta)+\cos(\theta)$ is just the sum of the two legs, which is greater than the length of the hypotenuse.
A: 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 $1$.
A: here is way to see this with a little bit of geometry. let $$x = \cos t, y = \sin t.$$  therefore  $$ \cos t + \sin t = x + y$$ to find the value of $$\cos t + sin t, 0 \le t \le \pi/2,$$ is the value $k$ as the $$ x + y = k  $$ cuts the unit circle in the first quadrant.  we know that for $k = 1,$ the  line cuts at $(1, 0)$ and $(0,1).$ because the unit circle is convex, the value of $k \ge 1.$   that is $$\cos t + \sin t \ge 1 \text{ for } 0 \le t \le \pi/2.$$
