# 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?

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• In the inductive step $$(n+1-1)+(\frac{b}{a})^{n+1} \color{red}{\geq} (n+1)\cdot \frac{b}{a}$$, you have already assumed the inequality that you have to prove to be true, which is incorrect. – Anurag A Aug 14 at 9:09
• Do you really have to prove this by induction, or any proof is allowed? This is a kind of the AM-GM inequality. – szw1710 Aug 14 at 9:09
• You did not prove anything. You proved something like $P_{n+1}$ iff $P_{n+1}$! – Kavi Rama Murthy Aug 14 at 9:09
• @szw1710 Yes, induction is mandatory – Matthias W Aug 14 at 9:10
• @AnuragA you are right. Hate those inequalities... – Matthias W Aug 14 at 9:12

## 1 Answer

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:)