It is very easy using calculus: Let $f(a,b)=a^b-b(a+1)+1$ So we need to prove $f(a,b)\ge 0$ for $a\ge 2$ and $b\ge 3$ $\frac{\partial{}f(a,b)}{\partial{}a}=ba^{b-1}-b>0$ We need to prove just $f(2,b)\ge0$. But $\frac{\partial{}f(2,b)}{\partial{}b}=2^b\cdot ln(2)-3 >0$ So $f(a,b)\ge f(2,3)=0$

MathUniverse wrote: $\frac{\partial f(2,b)}{\partial b}=2^b\cdot\ln 2 -3>0$ To be more strict, the inequality $2^b\cdot\ln 2-3>0$ when $b\ge 3$ i.e. $8\ln 2-3>0$ should be proved additionally. According to my own inequality $\ln 2>\frac{2}{3}$ established http://www.artofproblemsolving.com/Forum/viewtopic.php?f=296&t=351267, we have $\ln 2>\frac{2}{3}>\frac{3}{8}$, which ends the proof.

WakeUp wrote: Let $a$ and $b$ be positive integers with $a>1$ and $b>2$. Prove that $a^b+1\ge b(a+1)$ and determine when there is inequality. I think it can be done by induction, but it's a bit complicated

Since $a,b\in N\implies a\geq2, b\geq3$ $ \frac{a^b-1}{a-1}=1+a+a^2+a^3+....+a^{b-1}\geq b a^{\frac{b-1}{2}}$ But $b\geq3\implies \frac{b-1}{2}\geq 1\implies \frac{a^b-1}{a-1}\geq ab$ For the values of $a\geq3, b\geq3$ $a^b+1\geq ab(a-1)+2\implies \frac{a^b+1}{b}\geq a(a-1)+\frac{2}{b}>a(a-1)>a+1$ $a^b+1>b(a+1)$ for $a\geq3, b\geq3$ For $a=2$ we have to prove that $2^b+1\geq3b$ which is true for $b\geq3$