Q:

Showing that a certain inequality holds for all $ x \in \mathbb{R} $ and $ n \in \mathbb{N} $.

I would appreciate it if somebody could help me with the following problem.

Show that
  $$
\forall x \in \mathbb{R}, ~ \forall n \in \mathbb{N}: \quad
(x + 1)^{2n + 1} - x^{2n + 1} \geq 2 \left( \frac{1}{2} \right)^{2n + 1}.
$$

A:

When $x \geq 0$, we have $(x+1)^{2n+1}-x^{2n+1}=(1+(2n+1)x+ \ldots +(2n+1)x^{2n}) \geq 1 \geq 2 \left(\frac{1}{2}\right)^{2n+1}$.
When $x \leq -1$, take $y=-(x+1) \geq 0$, so $(x+1)^{2n+1}-x^{2n+1}=(y+1)^{2n+1}-y^{2n+1} \geq 2 \left(\frac{1}{2}\right)^{2n+1}$ by above.
When $-1<x<0$, take $y=-x, 0<y<1$, so $(x+1)^{2n+1}-x^{2n+1}=y^{2n+1}+(1-y)^{2n+1} \geq 2\left(\frac{y+(1-y)}{2}\right)^{2n+1}=2 \left(\frac{1}{2}\right)^{2n+1}$ by Power mean inequality.

