Simple doubt on a linear algebra proof involving inequalities

Question

From Lang's Introduction to Linear Algebra (1985), page 27:

Theorem 4.2: Let $A, B$ be two vectors in $R^n$. Then $|A·B|\le||A||\ ||B|| $.
Proof: Let $c=(A·B)/(B·B)$. We write $$A=A-cB+cB$$ By Pythagoras, $$||A||^2=||A-cB||^2+||cB||^2=||A-cB||^2+c^2||B||^2 $$ Hence $c^2||B||^2\le||A||^2$ (and the proof continues)

My doubt is, where exactly does $c^2||B||^2\le||A||^2$ come from? Shouldn't the sign be $\ge$, since $||A||^2=||A-cB||^2+c^2||B||^2$ and $||A-cB||^2$ is always greater than or equal to zero, so that $||A||^2\le c^2||B||^2$? Thanks.

Answer 1

Both $\|A-cB\|^2$ and $c^2\|B\|^2$ are nonnegative and they add up to $\|A\|^2$. Therefore none of them is greater than $\|A\|^2$ itself and so each quantity must be at most $\|A\|^2$.

Answer 2

We have $$\|A\|^2 = \|A - cB\|^2 + c^2 \|B\|^2 \geq 0 + c^2\|B\|^2$$ so indeed, $\|A\|^2 \geq c^2\|B\|^2$.