 Solution

For the Cauchy-Schwarz inequality, take $\lambda =(\mathbf{x}\cdot \mathbf{y})/(\mathbf{y}\cdot \mathbf{y})$ and use positivity of the inner product:Multiplication by $\mathbf{y}\cdot \mathbf{y}>0$ provesEquality holds exactly when $|\mathbf{x}-\lambda \mathbf{y}{|}^2=0$. Positive definiteness of the inner product makes this equivalent to $\mathbf{x}=\lambda \mathbf{y}$. Conversely, substitution of a scalar multiple gives equality. Thus equality holds precisely for linearly dependent nonzero vectors.