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Because both sides are $\geq 0$, we square them, and we get $$ \sum\limits^n_{k=1}(x_k-y_k)^2 \leq \sum\limits^n_{k=1}x_k^2 + \sum\limits^n_{k=1}y_k^2+2\sqrt{\sum\limits^n_{k=1}x_k^2\sum\limits^n_{k=1}y_k^2} $$ $$ \sum\limits^n_{k=1}x_k^2+\sum\limits^n_{k=1}y_k^2-2\sum\limits^n_{k=1}x_ky_k\leq \sum\limits^n_{k=1}x_k^2 + \sum\limits^n_{k=1}y_k^2+2\sqrt{\sum\limits^n_{k=1}x_k^2\sum\limits^n_{k=1}y_k^2} $$ $$ -2\sum\limits^n_{k=1}x_ky_k\leq 2\sqrt{\sum\limits^n_{k=1}x_k^2\sum\limits^n_{k=1}y_k^2}$$ I'm not sure how to prove this.

Gjekaks
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2 Answers2

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If we think geometrically, we are given $x,y\in\mathbb{R}^n$, and we want to show that $$\|x-y\|\leq \|x\| + \|y\|$$ which follows readily from the triangle inequality: we have $$ \|x\| + \|y\| = \|x\| + \|-y\| \geq \|x-y\|$$

Edit: as one of the comments quite rightly pointed out, it should have been obvious to me from the beginning you were asking about a proof of the triangle inequality, so sorry about that. In the above notation, you've reduced the question to proving that $-\langle x, y\rangle \leq \|x\|\|y\|$, equivalent to $\langle x,y\rangle \leq \|x\|\|y\|$ after say multiplying $x$ by $-1$. This is precisely the content of the Cauchy-Schwarz inequality, which has been discussed on this forum, in this question.

Community
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amakelov
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I'll give the answer in terms of the norm and you can work out how to transform it into whatever argument form you require. First note that $$\|u-v\|^2=(u-v)\cdot (u-v)=u\cdot u-2u\cdot v -v\cdot v=\|u\|^2-2u\cdot v+\|v\|^2$$ Then note that $$(\|u\|+\|v\|)^2=\|u\|^2+2\|u\|\|v\|+\|v\|^2$$ Finally it can be seen with the Cauchy Schwarz inequality (proof readily available with a quick Google search) that $$-2u\cdot v=2(-u)\cdot v\leq 2\|-u\|\|v\|=2\|u\|\|v\|$$ So in conclusion $$\|u-v\|^2\leq (\|u\|+\|v\|)^2$$ As $t^2$ is monotone increasing on the positive axis the proof follows.

K.Power
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