Suppose $a,b$ are real numbers and $0 < a < b$. Prove $ba^n + ab^n < a^{n+1} + b^{n+1}$

Question

$a,b$ are real numbers and $0 < a < b$

Prove that for all $n ≥ 1$, $ba^n + ab^n < a^{n+1} + b^{n+1}$

My attempt:

By contradiction. Suppose

$$\tag{* }ba^n + ab^n ≥ a^{n+1} + b^{n+1}$$

Rearranging $(*)$ gives

$$\begin{align} 0 & ≥ a^{n+1} + b^{n+1} - ba^n - ab^n \\ & ≥ a^{n+1} - ba^n + b^{n+1} - ab^n \\ & ≥ a^n(a - b) - b^{n}(a - b) \\ & ≥ (a - b)(a^n - b^n) \end{align}$$

But since $0 < a < b$, we have $(a-b) < 0$ and $(a^n - b^n) < 0$, and thus $(a-b)(a^n - b^n) > 0$.

Hence a contradiction. $\Box$

1. Is it correct?

2. Is there a better way? (Preferably direct proof)

Answer 1

Yes, it is correct. But it is more natural to apply the same idea to get a direct proof:$$a^{n+1}+b^{n+1}-ba^n-ab^n=(b^n-a^n)(b-a)>0.$$

Answer 2

According to the Rearrangement inequality: $$0<a<b \Rightarrow 0<a^n<b^n\\ ba^n + ab^n < b\cdot b^n +a\cdot a^n=a^{n+1} + b^{n+1}.$$