2
$\begingroup$

I have a statement which I'd have needed to solve a textbook exercice. I found a way around it, but I'm still curious about it - it seems reasonable, but I can't find the right argument for it. Here it is.

Let $a_1, \ldots, a_n, b_1, \ldots, b_n$ be non-negative real numbers. If $$ \sum_{i = 1}^n a_i b_i = \sum_{i = 1}^n a_i^2 = \sum_{i = 1}^n b_i^2 $$

then $a_i = b_i$ for each $1 \leq i \leq n$.

Also, does that work if we drop the second equality?

$\endgroup$
12
$\begingroup$

Consider the sum $$\sum_i (a_i - b_i)^2 = \sum_i a_i^2 - 2 \sum_i a_ib_i + \sum b_i^2 = 0$$

If the sum of non negative numbers is equal to $0$ then each one is singularly equal to $0$, and this implies $$a_i = b_i $$ for every $i$

Note that you don't need any assumption on the sign of $a_i$, $b_i$

If you drop the second equality it's still true if you have $a_i > 0$, $a_i \le b_i$ as you can check looking at the sum $$\sum _i a_i(b_i - a_i)=0$$

But in general it's not true; for example take $a_1 = a_2 = 1$, $b_1 = 2, b_2 = 0$.

$\endgroup$
  • $\begingroup$ Great answer! Thanks a lot. $\endgroup$ – Manuel Lafond Aug 15 '16 at 13:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.