Homepage › Solution manuals › Gilbert Strang › Introduction to Linear Algebra › Exercise 1.2.22
v12w12 + 2v1w1v2w2 + v22w22 ≤ v12w12 + v12w22 + v22w12 + v22w22 is true (cancel 4 terms) because the difference is v12w22 + v22w12 − 2v1w1v2w2 which is (v1w2 − v2w1) 2 ≥ 0