<s>[INST] Proving $ab^n + ba^n < a^{n+1} + b^{n+1}$ by Induction 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
 [/INST] 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
$$n > 0.$$
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