Show that |A+ B| |A - B| < | A|^2 + | B |^2
To prove this statement, we'll use the triangle inequality:
|A+ B| ≤ |A| + |B|
and
|A - B| ≤ |A| + |B|
Now, multiply these two inequalities:
(|A+ B|)(|A - B|) ≤ (|A| + |B|)^2
Expanding the right side:
|A+ B||A - B| ≤ |A|^2 + 2|A||B| + |B|^2
Next, we'll show that:
|A|^2 + 2|A||B| + |B|^2 = |A|^2 + |B|^2
Consider the inequality:
2|A||B| ≤ 2|A||B| + |A|^2 + |B|^2
This is true because both |A|^2 and |B|^2 are non-negative values, so adding them to the original inequality does not make it false.
Therefore:
|A|^2 + 2|A||B| + |B|^2 ≤ |A|^2 + |B|^2
Substituting this into the previous inequality:
(|A+ B|)(|A - B|) ≤ |A|^2 + |B|^2
Thus, we have shown that:
|A+ B| |A - B| ≤ |A|^2 + |B|^2
Hence, we have proved the statement.
To show that |A + B| |A - B| < |A|^2 + |B|^2, we will use a direct proof.
Let A and B be vectors in an arbitrary vector space.
First, we expand the left side of the inequality:
|A + B| |A - B| = ||A|| |A - B| (since A + B = A implies B = 0, and we know that the magnitude of the 0 vector is 0.)
= ||A|| ||A - B|| (since |A| means the magnitude of the vector A)
= ||A|| ||A - B|| ||A|| (dividing both sides by ||A||, which is nonzero since A ≠ 0)
Now, using the triangle inequality, we can write:
||A - B|| ≤ ||A|| + ||B|| (since ||A - B|| = ||A + (-B)|| ≤ ||A|| + ||-B|| = ||A|| + ||B||)
Substituting this inequality back into the equation, we have:
||A|| ||A - B|| ||A|| ≤ ||A|| {||A|| + ||B||}
Expanding the right side, we get:
||A|| ||A - B|| ||A|| ≤ ||A||^2 + ||A|| ||B||
Finally, noting that ||A|| ||B|| ≤ ||A||^2 + ||B||^2 (which can be proven separately), we can write:
||A|| ||A - B|| ||A|| ≤ ||A||^2 + ||A|| ||B|| ≤ ||A||^2 + ||B||^2
Therefore, we have shown that |A + B| |A - B| < |A|^2 + |B|^2.