<s>[INST] Proving inequality by induction produces an unclear answer 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. 
 [/INST] 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}
$
</s>