<s>[INST] Suppose $a,b$ are real numbers and $0 < a < b$. Prove $ba^n + ab^n < a^{n+1} + b^{n+1}$ 
$a,b$ 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. $\Box$


*

*Is it correct?

*Is there a better way? (Preferably direct proof)
 [/INST] According to the Rearrangement inequality:
$$0<a<b \Rightarrow 0<a^n<b^n\\
ba^n + ab^n < b\cdot b^n +a\cdot a^n=a^{n+1} + b^{n+1}.$$
</s>