2
$\begingroup$

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?

New contributor
Matthias W is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Anurag A Aug 14 at 9:09
  • $\begingroup$ Do you really have to prove this by induction, or any proof is allowed? This is a kind of the AM-GM inequality. $\endgroup$ – szw1710 Aug 14 at 9:09
  • 1
    $\begingroup$ You did not prove anything. You proved something like $P_{n+1}$ iff $P_{n+1}$! $\endgroup$ – Kavi Rama Murthy Aug 14 at 9:09
  • $\begingroup$ @szw1710 Yes, induction is mandatory $\endgroup$ – Matthias W Aug 14 at 9:10
  • $\begingroup$ @AnuragA you are right. Hate those inequalities... $\endgroup$ – Matthias W Aug 14 at 9:12
1
$\begingroup$

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:)

$\endgroup$

Your Answer

Matthias W is a new contributor. Be nice, and check out our Code of Conduct.

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.