Prove that $\sin(x) + \cos(x) \geq 1$ $\forall x\in[0,\pi/2]: \sin{x}+\cos{x} \ge 1.$
I am really bad at trigonometric functions, how could I prove it?

Re-arrange to get $\cos(x) < 1 - \sin(x)$
We know that $\sin^2 (x) + \cos^2(x) = 1$ and therefor that $|\cos(x)| = \sqrt{1-\sin^2(x)}$
Now we want to know if $\sqrt{1-\sin^2(x)} < 1 - \sin(x)$
Now we square both sides: $1-\sin^2(x) < (1 - \sin(x))^2 = 1 - 2\sin(x) + \sin^2(x)$
Re-arranging again we get $0 < 2\sin^2(x) - 2\sin(x)$
In order for $\sin^2(x) > \sin(x)$ we would need a $\sin(x)$ value greater than 1. This is not possible, and therefor neither is the original form of the inequality $\cos(x) < 1 - \sin(x)$.
