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Let $x_1,\ldots,x_n$ be real numbers such that $|x_1|\geq \ldots\geq |x_n|$ and $\displaystyle \sum_{i=k+1}^n |x_i| \leq \alpha \sum_{i=1}^k |x_i|$ where $1\leq k \leq n-1$ and $\alpha >0$.

Prove that $$\sum_{i=k+1}^n x_i^2 \leq \alpha \sum_{i=1}^k x_i^2$$

I have managed to prove a looser inequality: $$\begin{align} \alpha \sum_{i=1}^k x_i^2 &\geq \frac{\alpha}{k}\left(\sum_{i=1}^k |x_i| \right)^2 \quad \quad \text{Cauchy-Schwarz}\\ &\geq \frac 1k \sum_{i=k+1}^n |x_i| \sum_{i=1}^k |x_i|\\ &\geq \frac 1k \left(\sum_{i=k+1}^n |x_i| \right)^2\\ &\geq \frac 1k \sum_{i=k+1}^n x_i^2 \quad \quad \text{since } \|z\|_1 \geq \|z\|_2 \end{align}$$

Since $\frac 1k$ is not $\geq 1$, I'm stuck...

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It is quite simple: Note that $|x_i| \le |x_k|$ for any $i \in \{k+1,\ldots,n\}$. Thus $$\tag{1}\sum_{i=k+1}^n x_i^2 \le |x_k| \sum_{i=k+1}^n |x_i| \le \alpha |x_k| \sum_{i=1}^k |x_i|,$$ where the given condition is used in the last step. Now $|x_k x_i| \le x_i^2$ for an $i=1,\ldots,k$, because $|x_k| \le |x_i|$ for all $i=1,\ldots,k$. So proceeding in (1) gives the claimed estimate $$\sum_{i=k+1}^n x_i^2\le\alpha \sum_{i=1}^k x_i^2.$$

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  • $\begingroup$ Nice, thank you. This line of thought did not occur to me. $\endgroup$ – Issou Chankla Nov 21 '18 at 11:17
  • $\begingroup$ In general you lose information, if you apply Cauchy-Schwarz. The constants between $\|\cdot\|_1$ and $\|\cdot\|_2$ are also strict. In fact, the key point here is the condition $|x_1| \ge \ldots \ge |x_n|$. (The statement is wrong without this condition!) $\endgroup$ – p4sch Nov 21 '18 at 12:04

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