Prove $x^n + x < (x^n)x$ using induction.

Question

I need to prove $x^n$ + x < $x^n\cdot$ x, n $\in$ N, x $\in$ R>2 using induction. I started by

$x^n$ + x + (x^(n+1)+x) < ($x^n\cdot$ x) + (x^(n+1)+x)

I simplified to this:

< 2x^(n+1) + x

I am stuck here. All help is appreciated.

Answer 1

Let $P(n): x^n+x< x^{n+1}, n \in \mathbb{N}, x > 2$.

Check $P(1)$ is true. $x^1+x = 2x < x^2 \iff x(x-2) > 0$ is true since $x > 2$.

Assume $P(n)$ is true, i.e. $x^n + x < x^{n+1}$, prove $P(n+1)$ is true.

$x^{n+1} + x = x\cdot x^n + x < x(x^{n+1}-x) + x = x^{n+2} - x^2 + x < x^{n+2} - x^2 + x^2 = x^{n+2}$ since $x^2 > x$. Thus $P(n+1)$ is true, and by MPI the statement is true.