Prove that, for every natural number $n$, $x^n +n ≥ nx + 1$ for all $x>0$

Question

Prove that, for every natural number $n$, $x^n +n ≥ nx + 1$ for all $x>0$

This is the first part of the question, and I am not too sure how to prove it. My steps to prove it were to find a maximum-minimum case for $x$ and $n$ (ie. when $x → ∞$ and $n → ∞$, but I was not able to find a solution from there. Any help for this is appreciated.

P.S. I also need to prove that this inequality holds for $n= 3/2$, but I think that this can be done if I gain an understanding first of the above proof, so no spoilers please for proving the $n=3/2$ part $:)$

Answer 1

Using Bernoulli's Inequality: $$x^n=(1+(x-1))^n \geqslant 1+n(x-1)=nx+1-n$$