I need to prove following assertion by induction: $∀n ∈ N , ∀ a,b >0: a^{n-1}b \leq (\frac{(n-1)a+b}{n})^n$
After successfully proving the assertion for n = 1, I establish the following assumption: $a^{k-1}b \leq (\frac{(k-1)a+b}{k})^k$
Now when I try to prove for k + 1, which yields $a^{k}b \leq (\frac{((k+1)-1)a+b}{k+1})^{k+1}$, I get following inequality: $a^{k-1}a b \leq ((\frac{(k-1)a+b}{k})^{n} ) a $.
and this is where I have some troubles. I am unable to proof that $((\frac{(k-1)a+b}{k})^{n} ) a \leq (\frac{((k+1)-1)a+b}{k+1})^{k+1}$, which ultimately hinders me from finding a solution.
The inequality is $a^{n-1}b \leq (\frac{(n-1)a+b}{n})^n = (a+\frac{b-a}{n})^n = a^n (1+\frac{b/a-1}{n})^n $ or $b/a \le (1+\frac{b/a-1}{n})^n $ or, letting $r = b/a$, $r \le (1+(r-1)/n)^n $ or, letting $r = 1+ns$, $1+ns \le (1+s)^n $.
This is Bernoulli's inequality which is easily proved by induction. The induction step is
$\begin{array}\\ (1+s)^{n+1} &=(1+s)^n(1+s)\\ &\ge (1+ns)(1+s)\\ &= 1+(n+1)s+ns^2\\ &\ge 1+(n+1)s\\ \end{array} $