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So I know that $\sum x_n^2$ converges, and $\sum y_n^2$ converges. ${x_n}$ and ${y_n}$ are both sequences of real numbers. How can I prove that $$\sum(x_n+y_n)^2$$ converges? So far I have $$\sum(x_n+y_n)^2=\sum(x_n)^2 +\sum (y_n)^2 +\sum(2x_ny_n)$$ so I need to prove that $\sum 2x_ny_n$ converges. Is this the right way to approach it? If it is, I am completely lost as to the next step.

I also need to prove that $$\sqrt{\sum(x_n+y_n)^2}\leq \sqrt{\sum(x_n)^2}+\sqrt{\sum(y_n)^2}$$

I assume this will follow pretty easily from a result in the first proof, but I'm stuck on that now, and perhaps it is easier to start with the second one. I have a lot of work I've done on the first problem, but none of it seemed to go anywhere. Any hints?

Thanks!

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  • $\begingroup$ The second result is called Minkowski inequality. Look at the proof here for $L^p$ spaces and try adapt it to $\ell^2$: en.wikipedia.org/wiki/Minkowski_inequality In this case, Holder is just Cauchy-Schwarz. $\endgroup$
    – Julien
    Feb 21, 2013 at 2:51
  • $\begingroup$ Thanks, that's just the lead I needed! $\endgroup$
    – JohnDvorak
    Feb 21, 2013 at 3:05
  • $\begingroup$ $$0\le \sqrt{xy} \implies x+y\le x+2\sqrt{xy}+y\implies \sqrt{x+y}\le \sqrt{x}+\sqrt{y}$$ $\endgroup$
    – Bumblebee
    Nov 15, 2016 at 20:01

3 Answers 3

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$$x^2 + 2xy + y^2 \geq 0$$

Similarly,

$$ x^2 - 2xy + y^2 \geq 0$$

Therefore

$$ -x^2 - y^2 \leq 2xy \leq x^2 + y^2 $$

So

$$ |2xy| \leq x^2 + y^2 $$

This shows that $s_n = \displaystyle\sum_{i=1}^n |2x_i y_i|$ is bounded and increasing sequence and so is convergent. So the series $\displaystyle\sum_i 2x_iy_i$ is absolutely convergent and therefore must be convergent.

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  • $\begingroup$ This is very helpful, thank you! How did you get from line 1 to line 2? That's the step I'm not getting. $\endgroup$
    – JohnDvorak
    Feb 21, 2013 at 2:08
  • $\begingroup$ The two first lines follow from $u^2\geq 0$ by letting $u=x+y$ and $u=x-y$ respectively @SylarMorgan $\endgroup$
    – Pedro
    Feb 21, 2013 at 2:10
  • $\begingroup$ @PeterTamaroff I see now, thanks! $\endgroup$
    – JohnDvorak
    Feb 21, 2013 at 2:11
  • $\begingroup$ +1 Very beautiful answer and explanation sir. $\endgroup$ Mar 4, 2019 at 6:22
  • $\begingroup$ Thanks, simpler more logical answer though: 0 <= |x + y| <= |x|+|y| <= 2max(|x|, |y|). So |x+y|^2 <= 4max(|x|, |y|)^2 <= 4x^2 + 4y^2 $\endgroup$
    – muzzlator
    May 31, 2019 at 9:44
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Use Cauchy-Schwarz, it will yield the result immediately.

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  • $\begingroup$ For the second proof, you mean? $\endgroup$
    – JohnDvorak
    Feb 21, 2013 at 2:13
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$2x_n^2 + 2y_n^2 - (x_n + y_n)^2 $ $= 2x_n^2 + $ $2y_n^2 - x_n^2 -$ $ 2x_n y_n - y_n^2 = (x_n - y_n)^2 \geq 0$.

This means $(x_n + y_n)^2 \leq 2x_n^2 + 2y_n^2$.

Now add in $n$ and you see $\sum_n (x_n + y_n)^2$ converges.

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  • $\begingroup$ Your algebraic approach made it very easy to understand, thanks! $\endgroup$
    – JohnDvorak
    Feb 21, 2013 at 2:33

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