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It is not clear for me how the triangle inequality is used here:

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And by the triangle inequality (1.3) $$\left(\sum\limits_{i=1}^n |\xi_i+\eta_i|^2\right)^{1/2} \le \left(\sum\limits_{i=1}^n |\xi_i|^2\right)^{1/2} + \left(\sum\limits_{i=1}^n |\eta_i|^2\right)^{1/2}.$$

Especially because the absolute value on the left side is squared and we have taken the square root of all the summation on the left side.could anyone clarify this for me?

Edit:

I think I should have corrected my above question "How to prove Minkowski inequality"?

Thanks!

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    $\begingroup$ How would you express the distance between two points in Euclidean space using co-ordinates? (apply Pythagoras). Here you are comparing one distance on the left with the sum of two distances on the right. It is just the co-ordinate version of the triangle inequality. $\endgroup$ – Mark Bennet Oct 16 '17 at 7:37
  • $\begingroup$ I think I should have corrected my above question to >>> How to prove Minkovisky inequality ?>>>>> is my question now clear for you?@MarkBennet $\endgroup$ – Intuition Oct 16 '17 at 7:53
  • $\begingroup$ @MarkBennet yes you are right .....Thank you I have known the answer ..... it is just by substituting the definition of the norm in the triangle inequality. $\endgroup$ – Intuition Oct 16 '17 at 9:47
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Hint. By squaring both sides we obtain the equivalent inequality $$\sum_{i=1}^n |\xi_i+\eta_i|^2\leq \left(\left(\sum_{i=1}^n |\xi_i|^2\right)^{1/2}+\left(\sum_{i=1}^n |\eta_i|^2\right)^{1/2}\right)^2$$ that is $$\sum_{i=1}^n |\xi_i|^2+\sum_{i=1}^n |\eta_i|^2+2\sum_{i=1}^n |\xi_i||\eta_i|\leq \sum_{i=1}^n |\xi_i|^2+\sum_{i=1}^n |\eta_i|^2+2\sqrt{\sum_{i=1}^n |\xi_i|^2}\sqrt{\sum_{i=1}^n |\eta_i|^2}.$$ Can you take it from here?

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  • $\begingroup$ But where is the use of the triangle inequality in what you said? $\endgroup$ – Intuition Oct 16 '17 at 7:08
  • $\begingroup$ In your book, what does (1.3) say? $\endgroup$ – Robert Z Oct 16 '17 at 7:17
  • $\begingroup$ I think I should have corrected my above question to >>> How to prove Minkovisky inequality ?>>>>> is my question now clear for you? $\endgroup$ – Intuition Oct 16 '17 at 7:51
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    $\begingroup$ @Intuition See my edited answer. $\endgroup$ – Robert Z Oct 16 '17 at 8:12
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    $\begingroup$ @Intuition No, By squaring we obtain an inequality equivalent to the original one. So my proof is showing that your inequality is equivalent to Cauchy-Schwartz inequality (that you already know). $\endgroup$ – Robert Z Oct 16 '17 at 9:47
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This is the triangle inequality for the Euclidean norm: for the vector $v=(\xi_1,\dots,\xi_n)\in\mathbb{C}^n$, $$ \|v\|=\biggl(\,\sum_{i=1}^n|\xi_i|^2\biggr)^{\!1/2} \tag{*} $$ Since this is a norm, it satisfies the triangle inequality $$ \|v+w\|\le\|v\|+\|w\| $$ You should be able to find the proof about (*) defining a norm in the book you're reading or any other one on basic linear algebra.

Anyway, here's a proof that's based on the Cauchy-Schwarz inequality. The only difficult part is indeed the triangle inequality. On the other hand, the Cauchy-Schwarz inequality tells you that, for $v=(\xi_1,\dots,\xi_n)\in\mathbb{C}^n$ and $w=(\eta_1,\dots,\eta_n)\in\mathbb{C}^n$ $$ \biggl|\,\sum_{i=1}^n \overline{\xi_i}\eta_i\,\biggr|\le \biggl(\,\sum_{i=1}^n|\xi_i|^2\biggr)^{\!1/2} \biggl(\,\sum_{i=1}^n|\eta_i|^2\biggr)^{\!1/2} \tag{**} $$ Let's set $$ \langle v,w\rangle=\biggl|\,\sum_{i=1}^n \overline{\xi_i}\eta_i\,\biggr| $$ for simplicity, so $\langle w,v\rangle=\overline{\langle v,w\rangle}$ and, moreover, (**) reads $$ |\langle v,w\rangle|\le\sqrt{\langle v,v\rangle\langle w,w\rangle} \tag{***} $$ Now \begin{align} \|v+w\|^2 &=\langle v,v\rangle+\langle v,w\rangle+\langle w,v\rangle+\langle w,w\rangle \\[6px] &=\langle v,v\rangle+\langle v,w\rangle+\overline{\langle v,w\rangle}+\langle w,w\rangle \\[6px] &=\langle v,v\rangle+2\operatorname{Re}(\langle v,w\rangle)+\langle w,w\rangle\\[6px] &\le\langle v,v\rangle+2|\langle v,w\rangle|+\langle w,w\rangle &&\text{because $\operatorname{Re}(z)\le|z|$}\\[6px] &\le\langle v,v\rangle+\sqrt{\langle v,v\rangle\langle w,w\rangle}+\langle w,w\rangle &&\text{because of (***)} \\[6px] &=\bigl(\sqrt{\langle v,v\rangle}+\sqrt{\langle w,w\rangle}\,\bigr)^2 \end{align} which yields the required inequality.

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  • $\begingroup$ I think I should have corrected my above question to >>> How to prove Minkovisky inequality ?>>>>> is my question now clear for you? $\endgroup$ – Intuition Oct 16 '17 at 7:50
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    $\begingroup$ @Intuition I added the proof based on Cauchy-Schwarz $\endgroup$ – egreg Oct 16 '17 at 8:20
  • $\begingroup$ Thank you I have known the answer ..... it is just by substituting the definition of the norm in the triangle inequality. $\endgroup$ – Intuition Oct 16 '17 at 9:50
  • $\begingroup$ yeah ..... thank you so much your explanation is so clear. $\endgroup$ – Intuition Oct 16 '17 at 10:55

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