Show that |A+B| |A-B|≤ |A|2 +|B|2

Answer 1

Starting with left-hand side:

|A+B|² - |A-B|²

= (A+B)⋅(A+B) - (A-B)⋅(A-B) (using |x|² = x⋅x)

= A⋅A + B⋅B + 2A⋅B - A⋅A - B⋅B + 2A⋅B

= 4A⋅B

Now divide both sides by 4 and multiply by 4(|A|² + |B|²):

|A+B| |A-B| ≤ 4A⋅B ≤ 2(A²+B²) ≤ |A|² + |B|²

The second inequality follows from the fact that (A-B)² = A² - 2A⋅B + B² ≥ 0, which rearranges to 2A⋅B ≤ A²+B².