Proving $ab^n + ba^n < a^{n+1} + b^{n+1}$ by Induction

Question

Given $0<a<b , ab \in \mathbb{R}$ , $n \geq 1$

By induction, Assuming $ab^n + ba^n < a^{n+1} + b^{n+1}$. - $(1)$

To prove $ab^{n+1} + ba^{n+1} < a^{n+2} + b^{n+2}$

Multiplying (1) by ab we get

$a^2b^{n+1} + b^2a^{n+1} < a^{n+2}b + b^{n+2}a$

How to proceed ?

Thanks

What are $a,b,n?$ Why do you want to prove it by induction? Can you write a context - where does the problem come from? — user376343, May 03 '22 at 17:46
If the inequality $a^nb+ab^n < aa^n + bb^n$ is true (which is already answered below) for $0<a<b, ab \in \mathbb{R}, n<0$, why do you want to prove it by induction? If you multiply it with $ab$ which is constant and positive, the fact remains the same. It is like, $a>b \Leftrightarrow 2a>2b$ which does not make sense. — Eddy Piedad, May 03 '22 at 18:51
@EddyPiedad because it is question from chapter named mathematical induction — Sophie Clad, May 03 '22 at 19:06
I agree with the first comment. You should provide the context. Maybe you can cite your book reference in your question. And also, you still need to define $n$. Your assumption in (1) is only valid for $n<0$ as shown in the answer. — Eddy Piedad, May 03 '22 at 20:27
Does this answer your question? Suppose $a,b$ are real numbers and $0 < a < b$. Prove $ba^n + ab^n < a^{n+1} + b^{n+1}$ . Found using Approach0. — John Omielan, May 03 '22 at 21:10

Eddy Piedad · Accepted Answer · 2022-05-03T21:39:30.150

2

Your equation is equivalent to $$0 < aa^n-a^nb-ab^n+bb^n$$ $$0 < (a-b)(a^n-b^n)$$

Since $0 < a < b,$ where $ a,b \in \mathbb{R}$, $$0>(a-b)$$

that leaves $$0 > (a^n-b^n)$$

Working on $n$, we have $$(a^n-b^n) < 0$$ $$n\ln\frac{a}{b} < 0.$$

Again, $\ln\frac{a}{b}<0$ when $a<b$ so $$n > 0.$$

edited May 03 '22 at 21:39

answered May 03 '22 at 17:52

Eddy Piedad

It doesn't really matter $a<b$ – nonuser May 03 '22 at 17:54
Thanks but i would like to proceed from where i left if possible' – Sophie Clad May 03 '22 at 17:55
You should define $n$. And here you can see what the value of $n$ should be. – Eddy Piedad May 03 '22 at 18:37
@EddyPiedad yes i have edited. can it be solved from where i left or this is only way that you wrote ? – Sophie Clad May 03 '22 at 20:54
Yes @SophieClad. I have also edited and modified the typo. It should be true for all $n>0$. – Eddy Piedad May 03 '22 at 21:40

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