Proving that $\sqrt{a_1^2} +\cdots+ \sqrt{a_n^2} > \sqrt{a_1^2 +\cdots+a_n^2}$ using Pythagoras

Question

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.

Answer 1

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 )

Answer 2

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 $0$.

Answer 3

Here is a geometrical proof. Consider a square with side length $L = |a_1|+|a_2| + \ldots + |a_n|$.

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$

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.