<s>[INST] Prove that $a^{b} +1\geq b(a+1)$ $a, b \in \mathbb{N} $ and $a>1$,$b>2$
Prove that $a^{b} +1\geq b(a+1)$
My attempt :
suppose $a=p+1$, ($ p \in \mathbb{N}$,$ p\geq 2$)
Remarks :(if $ p=1$ hence $ a=2$,in this case isn't difficult)
So let's show $(p+1)^b +1\geq bp+2b$
We know that $(p+1)^b=\sum_{k=0}^{b}\left( {\begin{array}{c} b \\ k \\ \end{array}} \right) ×p^b\geq \left( {\begin{array}{c} b \\ 1 \\ \end{array}} \right) 
×p+\left( {\begin{array}{c} b \\ b \\ \end{array}} \right) ×p^b$
So $(p+1)^b+1\geq bp+p^b+1\geq bp+2b$
Because $p\geq 2$$\Rightarrow$$ p^b\geq 2^b\geq 2b$
Does my attempt is true?
 [/INST] We can prove this by induction on $b$ when $b\gt2 ,a\gt 1$.
Step 1:
$b=3$, $a^3+1\geq 3a+3$ is easy to prove, note that $a(a^2-3)\geq2$ since $a \geq2$.
Step2:
Now suppose we have $a^b+1\geq ab+b$.
$a^{b+1}+1=a(a^b+1)-a+1$
$ \geq a(ab+b)-a+1=
ab(a+1)-a+1\geq2b(a+1)-a+1$,
and since $b(a+1)-a+1 \gt 2(a+1)-a+1 \gt a+1$,we have $a^{b+1}+1\gt b(a+1)+a+1=(b+1)(a+1)$and we are done.
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