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I want to know if the following inequality holds: $(\sum\limits_{i=1}^{N}|x_i|-|y_i|)^{2}\le C\sum\limits_{i=1}^{N}(|x_i|-|y_i|)^{2}$ where $C>0$ is some constant.

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    Why don't you simply assume $x_i$, $y_i \ge0$? – 2017-01-29
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    Maybe I don't see it right be right side is equal to the left by only multiple by positive constant c. The inequality holds. – 2017-01-29
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    Is it allowed for $C$ to depend on $N$ ? If so, apply Schwarz inequality and you're done. – 2017-01-29
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    I think it has to depend on $n$. Cauchy Schwarz gives $c=n$, and the equality occurs, so $c=n$ is the best you can do. @Adren – 2017-01-29
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    Yes $C$ can depend on N. Is just wasn't sure if I could apply Cauchy Schwarz as we have a differene in the sum. – 2017-01-29

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Let's remind the Cauchy-Schwarz inequality in $\mathbb{R}^N$, for the standard inner product :

$$\forall a\in\mathbb{R}^N,\,\forall b\in\mathbb{R}^N,\,\left^2=\Vert a\Vert^2\Vert b\Vert^2$$

that is, more explicitly :

$$\forall (a_1,\cdots,a_N)\in\mathbb{R}^N,\,\forall(b_1,\cdots,b_N)\in\mathbb{R}^N,\,\left(\sum_{i=1}^Na_ib_i\right)^2\le\left(\sum_{i=1}^Na_i^2\right)\left(\sum_{i=1}^Nb_i^2\right)$$

Now if we choose $a_i=\vert x_i\vert-\vert y_i\vert$ and $b_i=1$ for every $i\in\{1,\cdots,N\}$, then we get :

$$\left(\sum_{i=1}^N\vert x_i\vert-\vert y_i\vert\right)^2\le N\sum_{i=1}^N\left(\vert x_i\vert-\vert y_i\vert\right)^2$$

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    Ok it all clear now. Thank you very much for your help. – 2017-01-29