For $x\in\mathbb{R}^n$, define $|x|^{\downarrow}$ as the vector of absolute values sorted, meaning $|x_1|\geq |x_2|\geq \dots\geq |x_{n-1}|\geq |x_n|$. I'd like to prove that: $$\langle x,y \rangle \leq \langle |x|^{\downarrow},|y|^{\downarrow} \rangle$$
So far I was able to prove this for a 2-dimensional vector and wanted to prove by induction, but wasn't successful.
Here is the proof so far: Assume $x_1\leq x_2$ and also $y_1\geq y_2$, and assume w.l.o.g that $x_1,x_2,y_1,y_2>0$. We want to show that $x_1y_1+x_2y_2\leq x_1y_2+x_2y_1$. Dividing by $x_1$ and rearranging the terms give us $y_1-y_2\leq \frac{x_2}{x_1}(y_1-y_2)$ and since we know that $y_1-y_2\geq 0$ and also $\frac{x_2}{x_1}\geq 1$ then we have proved $x_1y_1+x_2y_2\leq x_1y_2+x_2y_1$.
Any assistance on how to generalize this to n-dimensional vectors will be appreciated.
Best Answer
You have $$ \langle x,y \rangle \le |\langle x,y \rangle| \le \sum_{k=1}^n |x_k|| y_k| \leq \langle |x|^{\downarrow},|y|^{\downarrow} \rangle $$ where the last inequality is the rearrangement inequality.