I understand that with questions asking you to prove for x on an interval requires induction. The question I have is: prove that for every real number $x \in \left[0,\dfrac{\pi}{2}\right]$, $\sin(x)+\cos(x)\ge1$.
The base case is obvious: 0 satisfies the equation I want to prove.
The inductive case is more confusing. The hypothesis is that there is an n that satisfies the equation. But then what should I do for the n+1 case?
Would I do $n+2\pi$ because $2\pi$ is similar to adding 1?
Alt. hint: $\;(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2 \sin x \cos x = 1 + \sin 2x \ge 1\;$ (why?).
[ EDIT ] A complete proof would just need to piece together the following:
if $a \ge 0$ then $a \ge 1 \iff a^2 \ge 1$
if $x \in [0, \pi/2]$ then $\sin x \ge 0$ and $\cos x\ge 0$, so $\sin x + \cos x \ge 0$
therefore $\sin x + \cos x \ge 1 \iff (\sin x + \cos x)^2 \ge 1$ for $x \in [0,\pi/2]$
$x \in [0, \pi/2] \iff 2x \in [0, \pi]$, and $\sin$ is non-negative on $[0,\pi]$
it follows that $(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2 \sin x \cos x = 1 + \sin 2x \ge 1$