For every $x \in [\frac{\pi}{2},\pi]$, $\sin(x)+\cos(x)\geq 1$. Prove rigorously by contradiction. For every $x \in [o,\frac{\pi}{2}]$, $\sin(x)+\cos(x)\geq 1$.
How do you prove this rigorously by contradiction?

I understand you start by assuming that $\sin(x)+ \cos(x)&lt;1$ and prove this is a false statement. I can see from drawing the graphs this is false but how do I show it algebraically?
Both $\cos(x)$ and $-\sin(x)$ are convex functions on the interval $(\pi/2,\pi)$. Since in the endpoints of such interval se have $\cos(x)-\sin(x)=-1$, it follows that $\cos(x)-\sin(x)\leq -1$ all over $[\pi/2,\pi]$. The last inequality implies that $\sin(x)+\cos(x)$ is a decreasing function over $[\pi/2,\pi]$, so, for any $x\in(\pi/2,\pi]$,
$$\sin(x)+\cos(x)&lt;\sin(\pi/2)+\cos(\pi/2)=1.$$
We start with the assumption $$\sin x+\cos x&lt;1$$ and rearrange, $$\sin x&lt;1-\cos x$$ square both sides $$\sin ^2 x&lt;1-2\cos x+\cos^2x$$ and manipulate further. $$\sin^2x-\cos^2x&lt;1-2\cos x$$ Since $\sin^2x+\cos^2x=1$ and $\sin^2x+\cos^2x-2\cos^2x=\sin^2x-\cos^2x$, $$1-2\cos^2x&lt;1-2\cos x$$ $$2\cos^2x&gt;2\cos x$$ $$\cos x&gt;1$$ which is false since the codomain of $\cos x$ is $-1\le x\le 1$.