If $n\in N$, $n>1$ and $\alpha \in [0,1]$ prove that
$$(n^\alpha-(n-1)^\alpha)(2^\alpha+n-n^\alpha-(n-1)^\alpha-1)+(n-1)^\alpha \geq 1$$
EDIT: This is wrong, I misread my own translation as $\displaystyle \alpha \in \{0,1\}$, this problem remains unsolved
Case 1. $\displaystyle \alpha = 0$
$\displaystyle (n^0-(n-1)^0)(2^0+n-n^0-(n-1)^0-1)+(n-1)^0 \geq 1$
$\displaystyle (1-1)(1+n-1-1-1)+1 \geq 1$
$\displaystyle 1 \geq 1$
TRUE
Case 2. $\displaystyle \alpha = 1$
$\displaystyle (n^1-(n-1)^1)(2^1+n-n^1-(n-1)^1-1)+(n-1)^1 \geq 1$
$\displaystyle (n-(n-1))(2+n-n-(n-1)-1)+(n-1) \geq 1$
$\displaystyle (n-n+1)(2+n-n-n+1-1)+n-1 \geq 1$
$\displaystyle (1)(2-n)+n-1 \geq 1$
$\displaystyle 2-n+n-1 \geq 1$
$\displaystyle 1 \geq 1$
TRUE
Because Every value of alpha leads this to be true regardless of the value of n, this statement is always true.
Edit:
*sigh* stupid double posts >.<