Note $1/(1 - \alpha)$ cannot be the least upper bound when $-1 < \alpha < 0$. For if $\alpha = -1/2$, $1/(1 - \alpha) = 2/3$, but $1 + \alpha = 3/2 >...
Please attach the images inline...otherwise they won't show up in the "Topic Review" and having them show up there can be helpful when replying to a...
In order for three points to be collinear, we can pick any two distinct sets of two points from the set of three to form two line segments, and those...
Start with a bit of calculus. If $0<b\leqslant c$ and $f(x) = x(b+c-x)$, then $f(x) \geqslant bc$ whenever $b\leqslant x\leqslant c.$ The reason for...