Prove inequality $(n-1)\cdot a^n +b^n \geq n\cdot a^{n-1}\cdot b$ by induction let $a,b \in \mathbb{R}; a,b>0$
I have to prove the following inequality by induction:     
$(n-1)\cdot a^n +b^n \geq n\cdot a^{n-1}\cdot b$  with $n\in \mathbb{N}\setminus{\{1}\}$ 
I was not able to use the induction hypothesis for hours and of course did not reach the desired result: $n\cdot a^{n+1} +b^{n+1} \geq (n+1)\cdot a^n\cdot b$. 
Than I rearranged the inequality to
$(n-1)+(\frac{b}{a})^n \geq n\cdot \frac{b}{a}$ by dividing both sides by $a^n$ 
Doing induction for the rearranged inequality 
Base case: assumption is true for $n=2$
$(2-1)+(\frac{b}{a})^2 \geq 2\cdot \frac{b}{a} \Leftrightarrow a^2+b^2\geq 2ab \Leftrightarrow (a-b)^2\geq 0$ 
It remains to prove that the assumptions holds for $n+1$
This means $n\cdot a^{n+1} +b^{n+1} \overset{?}{\geq} (n+1)\cdot a^n\cdot b$ 
Inductive step:
$(n+1-1)+(\frac{b}{a})^{n+1} \geq (n+1)\cdot \frac{b}{a}$
$\Leftrightarrow n+(\frac{b}{a})^n\cdot(\frac{b}{a}) \geq n\cdot (\frac{b}{a})+(\frac{b}{a})$
$\Leftrightarrow \frac{n\cdot a}{b}+(\frac{b}{a})^n\geq n+1$
$\Leftrightarrow \frac{n\cdot a\cdot a^n+b^n\cdot b}{b\cdot a^n}\geq n+1$
$\Leftrightarrow \frac{n\cdot a^{n+1}+b^{n+1}}{b\cdot a^n}\geq n+1$
$\Leftrightarrow  n\cdot a^{n+1}+b^{n+1}\geq (n+1)\cdot a^n\cdot b$ 
So I reached the desired result, but did not use the induction hypothesis.
Can someone please point out the mistake(s) I have done? 

The problem in the inductive step is that, you started with the equation to be proved, and after solving it in a circular fashion, reached back where you started. The correct proof is as follows:
Since, $a$ and $b$ are positive, their ratio will be positive, let $x=\frac{b}{a}$. Now, what you want to show can equivalently be written as,
$$(n-1)+x^n\geq nx$$
as you also noted.
After rearranging,
$$x^n-nx+n-1\geq 0$$
Basis
For $n=2$, case it becomes $(x-1)^2\geq 0$.
Induction hypothesis
$$x^k-kx+k-1\geq 0$$
or equivalently,
$$x^k\geq kx-k+1$$
Inductive step
Consider $$x^{k+1}-(k+1)x+k$$
$$\geq (kx-k+1)x-(k+1)x+k$$
$$=k(x-1)^2\geq 0$$
Hence proved
Hope it helps:)
