Prove that for all $x$ where $01$ 
Prove that for all $x$ where $0<x<\pi/2$, $$\sin x + \cos x > 1.$$

I tried multiple Identities I do not know what I am missing. I have tried changing into different identities.
 A: For $x\in(0,\frac\pi 2)$ we have
$$\cos x+\sin x-1=\cos^2\left(\frac x2\right)-\sin^2\left(\frac x2\right)+2\sin\left(\frac x2\right)\cos\left(\frac x2\right)-1\\=-2\sin^2\left(\frac x2\right)+2\sin\left(\frac x2\right)\cos\left(\frac x2\right)=2\sin\left(\frac x2\right)\left(\underbrace{\cos\left(\frac x2\right)-\sin\left(\frac x2\right)}_{>0}\right)>0$$
A: For $0<x<\pi/2$, we have $0<\sin x <1$; thus 
$$\sin x>\sin^2 x,\text{ for }0<x<\pi/2.$$
Similarly,  
$$\cos x>\cos^2 x,\text{ for }0<x<\pi/2.$$
Thus, for $0<x<\pi/2$,
$$
\sin x+\cos x >\sin^2 x+\cos^2 x=1.
$$

You could also see why the result holds by considering a right triangle whose hypotenuse has unit length and appealing to the fact that the shortest distance between two points is a straight line. 
A: Another way which works by pure inspection of the sine and cosine functions would be to realize that
$$  \sin(x) > \frac{2x}{\pi}\quad\text{for}\quad 0<x<\pi/2,$$
i.e., the sine function is always above the straight line from the origin to the point $(\pi/2,1)$. The same is true for the cosine function:
$$\cos(x) > 1-\frac{2x}{\pi}\quad\text{for}\quad 0<x<\pi/2$$
Since the two lines add up to unity in the given interval, you have shown the inequality.
