Find all $x\in\mathbb R^{+}$, that satisfy this inequality $x^{n+1}-x^n-2 (-1)^n x\geq 0$ I would appreciate if somebody could help me with the following problem
Q: Find all $x\in\mathbb R^{+}$, that satisfy this inequality: (for all $n \in \mathbb{N}$)
$$x^{n+1}-x^n-2 (-1)^n x\geq 0 $$
 A: (Note: I'm assuming $0 \notin \mathbb{N}$, as otherwise there are trivially no solutions).
Let's rearrange. 
We need to find all $x$ satisfying the inequality $x^{n}(x-1) \ge 2(-1)^{n}x$ for every $n \in \mathbb{N}$.
First let's consider only $n$ even. Then the inequality becomes  $x^{n}(x-1) \ge 2x$. Solutions must be in $\mathbb{R}^{+}$, so we can rule out $x \le 1$, as this makes the left hand side negative. For $x > 1$ we have $x^{n+2}(x-1) \ge x^{n}(x-1)$, so if $x$ is a solution to the inequality for $n=2$, then it is a solution for all even $n$. So we need to solve $x^{3} - x^2 \ge 2x$. This gives $x(x+1)(x-2) \ge 0$, which is satisfied for $x \in \mathbb{R}^{+}$ precisely when $x \ge 2$.
Now let $n$ be odd. Then we have $x^{n}(x-1) \ge -2x$. But we must have $x \ge 2$ from the even case, so the left hand side is always positive, and then right hand side is always negative. So $x \ge 2$ is a solution for all odd $n$ as well.
Therefore the set of all $x \ge 2$ is the subset of $\mathbb{R}^{+}$ that satisfies the inequality for all $n$.
A: When $n$ is odd, we have
$$
x^{n+1}-x^n+2x \ge 0
\iff
x^{n}-x^{n-1}+2 \ge 0  
\iff x\ge 0
$$
Indeed, when $x\ge 1$, we have $x^{n}\ge x^{n-1}$.
When $0 \le x \le 1$, we have $x^{n}-x^{n-1}=x^{n-1}(x-1) \ge -1$
When $n$ is even, the solution is $x\ge a_n$, where $a_n\ge 1$, but which can only be given numerically for $n\ge 6$.
