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I think I have a proof using Pythagoras for $\sqrt{a_1^2} + \sqrt{a_2^2} > \sqrt{a_1^2 + a_2^2}$.

I'm interested in whether there's a way to use that proof with Pythagoras to prove the general $a_n$ case (for this, hints are appreciated rather than complete proofs), and also in other ways (algebraic, geometric, number theoric, calculusic...anything) that you might know or come up with to prove the general case (for those, either hints or complete proofs are great, up to you).

Lemma:

Let positive (edited) real numbers $a_1, a_2$ be the legs of a right triangle.

Then $\sqrt{a_1^2 + a_2^2}$ is the length of the hypothenuse of that triangle.

And $\sqrt{a_1^2} + \sqrt{a_2^2}$ is the sum of the length of the two legs.

By the triangular inequality, we know that the length of the hypothenuse has to be less than the length of the sum of the two legs.

Therefore, for any real numbers $a_1, a_2$, $\sqrt{a_1^2} + \sqrt{a_2^2} > \sqrt{a_1^2 + a_2^2}$.

I'm stuck here...I was thinking of comparing pairs of elements from each side of the expression using my lemma, but it doesn't seem possible to "extract" pairs of elements from under $\sqrt{a_1^2 + a_2^2 +...+a_n^2}$. I also thought about summing all elements but $a_1$ into a single number and using my lemma on those simplified expressions, but I run into the same problem.

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    $\begingroup$ Why do you need Pythagoras's theorem? Just square both sides of the inequality and compare. $\endgroup$ – Leo Jan 6 '16 at 19:18
  • $\begingroup$ When you say "by the triangular inequality", this is a bit circular - the identity you're trying to prove is the triangle inequality for the Euclidean distance. $\endgroup$ – πr8 Jan 6 '16 at 19:20
  • $\begingroup$ You have the inequality backwards. $\endgroup$ – zhw. Jan 6 '16 at 19:20
  • $\begingroup$ @zhw. It has been edited. $\endgroup$ – user236182 Jan 6 '16 at 19:26
  • $\begingroup$ @zhw. yes thank you, I mentioned it in my edit, not sure where it shows though $\endgroup$ – jeremy radcliff Jan 6 '16 at 19:27
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It's very easy using induction :

You proved the case when $n=2$ .

Now assume you know it for $n$ and want prove it for $n+1$ :

$$\sqrt{a_1^2}+\ldots+\sqrt{a_n^2}+\sqrt{a_{n+1}^2} >\sqrt{a_1^2+a_2^2+\ldots+a_n^2}+\sqrt{a_{n+1}^2}$$

Now use also the $n=2$ case to finnish it :

$$\sqrt{a_1^2+a_2^2+\ldots+a_n^2}+\sqrt{a_{n+1}^2}>\sqrt{\left ( \sqrt{a_1^2+a_2^2+\ldots+a_n^2} \right )^2}+\sqrt{a_{n+1}^2}>\sqrt{\left ( \sqrt{a_1^2+a_2^2+\ldots+a_n^2} \right )^2+a_{n+1}^2}=\sqrt{a_1^2+a_2^2+\ldots+a_n^2+a_{n+1}^2}$$ as wanted .

You can translate this into a geometric proof : consider an $n$-dimensional box with the sides $a_1,a_2,\ldots,a_n$ . The diagonal of the box is $\sqrt{a_1^2+a_2^2+\ldots+a_n^2}$ and now you can repeatedly apply the triangle's inequality to get your inequality (this is equivalent with the induction proof above )

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  • $\begingroup$ I like the geometric intuition you give a lot, it makes the algebra more meaningful to me. $\endgroup$ – jeremy radcliff Jan 6 '16 at 19:50
  • $\begingroup$ @jeremyradcliff But can you imagine an $n$-dimensional box (if $n\ge 4$)? $\endgroup$ – user236182 Jan 6 '16 at 20:41
  • $\begingroup$ @user236182, no of course, I'm stuck at 3 dimensions but the step of the diagonal of a 3d box gives me 2 steps (with the 2d case) to extrapolate about higher dimensions and makes the induction principle in this context a lot more meaningful to me. $\endgroup$ – jeremy radcliff Jan 6 '16 at 20:54
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Here is a geometrical proof. Consider a square with side length $L = |a_1|+|a_2| + \ldots + |a_n|$.

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$enter image description here

Comparing the area of the whole square to the area of the small squares contained within it we see that

$$(|a_1| + |a_2| + \ldots + |a_n|)^2 \geq a_1^2 + a_2^2 + \ldots + a_n^2$$

and by taking the square root we get the desired inequality. Equality can only happen when one of the small squares covers the whole square which can only happen when atleast $n-1$ of the $a_i$'s are zero.

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  • $\begingroup$ Very nice, thank you. I'm a big fan of proofs without words. $\endgroup$ – jeremy radcliff Jan 6 '16 at 19:54
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Both sides of the inequality are non-negative (for all $a_i\in\mathbb R$), therefore the following equivalences hold:

$$\sqrt{a_1^2} + \sqrt{a_2^2} +\cdots + \sqrt{a_n^2} \ge \sqrt{a_1^2 + a_2^2 +…+a_n^2}$$

$$\iff \left(\sqrt{a_1^2} + \sqrt{a_2^2} +\cdots + \sqrt{a_n^2}\right)^2 \ge \left(\sqrt{a_1^2 + a_2^2 +…+a_n^2}\right)^2$$

$$\iff a_1^2+a_2^2+\cdots+a_n^2+2\sum_{i=1}^n \sum_{j>i}^n|a_ia_j|\ge a_1^2+a_2^2+\cdots+a_n^2$$

$$\iff 2\sum_{i=1}^n \sum_{j>i}^n|a_ia_j|\ge 0,$$

which is true, with equality if and only if at least $n-1$ of $a_1,a_2,\ldots,a_n$ are equal to $0$.

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  • $\begingroup$ @jeremyradcliff It's multinomial expansion. $\endgroup$ – user236182 Jan 6 '16 at 19:58
  • $\begingroup$ Thank you. I erased my comment because I realized it was wrong. But I just worked it out. For the record, I was aking whether the summation symbols were from the binomial expansion. $\endgroup$ – jeremy radcliff Jan 6 '16 at 19:59

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