I use induction to prove that
$(1+x)^r \geqslant 1+rx$ when $r>1$ and $x\geqslant-1$
I have done the second step in the proof, but not the first one. Many solutions on the internet use r=1 because their $r$ was not bounded , but here $r>1$. I suppose I can use $r=2$, but then I can't see if the inequality is true or not.
As they said if you showed it for $r \ge 1$ then ofcourse it is true for $r>1$.
But if you want to prove it for $r=2$, you just simply do $(1+x)^2=1+2x+x^2 \ge 1+2x$ since $x^2\ge 0$ for any real number $x$.