For scalars a,b we know that (a-b)^2>=0=>2a^2+2b^2>=(a+b)^2 Is the analogous inequality for vectors also always true?

Tara Mayer

Tara Mayer

Answered question

2022-10-21

For scalars a,b we know that ( a b ) 2 0 2 a 2 + 2 b 2 ( a + b ) 2
Is the analogous inequality for vectors also always true?
i.e., does | | i a i | | 2 2 i | | a i | | 2 ?
Edit:
We can show similarly for 2 vectors that | | a + b | | 2 2 | | a | | 2 + 2 | | b | | 2
Now, for 3 vectors, we have | | a + b + c | | 2 2 | | a + b | | 2 + 2 | | c | | 2 4 | | a | | 2 + 4 | | b | | 2 + 2 | | c | | 2
Is there a general form? like | | i a i | | 2 α i | | a i | | 2

Answer & Explanation

rcampas4i

rcampas4i

Beginner2022-10-22Added 22 answers

What is true is i = 1 n a i 2 n i = 1 n a i 2 . This follows from triangle inequality for the norm and Cauchy-Schwarz inequality: i = 1 n a i i = 1 n a i ; now use the inequality ( i = 1 n ( 1 ) ( c i ) ) 2 ( i = 1 n 1 2 ) ( i = 1 n c i 2 ) valid for scalars c i

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?