Inequality. $(n+1)\left(a^{n+1}+b^{n+1}\right) \geq (a+b)\left(a^n+a^{n-1}b+\ldots+b^{n}\right). $ 
Let $a$ and $b$ be positive numbers, and $n \in \mathbb{N}$. Prove that (using Rearrangement Inequality)
$$(n+1)\left(a^{n+1}+b^{n+1}\right) \geq (a+b)\left(a^n+a^{n-1}b+\ldots+b^{n}\right). $$

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As the sequences $(a^{k+1},b^{k+1}),(a^{n-k},b^{n-k})$ are similarly sorted, therefore Rearrangement Inequality gives $a^{n+1} + b^{n+1} \ge a^{k+1}b^{n-k} + a^{n-k}b^{k}$.
The same inequality for $k-1$ is $a^{n+1} + b^{n+1} \ge a^{k}b^{n-k+1} + a^{n-k+1}b^{k-1}$, hence adding them gives $$a^{n+1} + b^{n+1} \ge (a+b)(a^{k}b^{n-k} + a^{n-k}b^{k})$$ Adding these inequalities for $k=0,1,2, \cdots, n $ gives the required.
