Help understand an inequality in a proof

The formula (10.26) says $$\langle (A-\lambda)V_ky,V_ky\rangle = \langle e_k,y\rangle \langle \hat{v}_{k+1},V_ky\rangle$$ Remembering (10.25), the angle between $(A-\lambda)V_ky$ and $V_ky$ is the same as the angle between $\hat{v}_{k+1}$ and $V_ky$, (the book does not say that this is fundamental for deriving (10.27)) so: $$\lVert(A-\lambda)V_ky\rVert \lVert V_ky\rVert = \langle e_k,y\rangle \lVert \hat{v}_{k+1}\rVert\lVert V_ky\rVert$$ For Cauchy-Schwartz applied to the first part of the rhs $$\langle e_k,y\rangle \le \lVert y\rVert$$

As to the lhs, remember that the 2-norm of a symmetric real-valued matrix $A$ is $$\lVert A\rVert=\max\limits_i \lvert \lambda_i\rvert$$ The 2-norm of its inverse, when A is non-singular, is then $$\lVert A^{-1}\rVert=1/\min\limits_i \lvert \lambda_i\rvert$$ So it is (this is an important bound formula, not mentioned in the book, rearranged for our purpose) $$\frac{\lVert(A-\lambda) V_ky\lVert}{\lVert V_ky\rVert} = \frac{\lVert(A-\lambda) V_ky\lVert}{\lVert (A-\lambda)^{-1}(A-\lambda)V_ky\rVert} \ge \frac{\lVert(A-\lambda) V_ky\lVert}{\lVert (A-\lambda)^{-1}\rVert\lVert(A-\lambda)V_ky\rVert} = \min\limits_i \lvert \lambda_i-\lambda\rvert$$ It now follows (10.27), that is $$\min\limits_i \lvert \lambda_i-\lambda\rvert \lVert V_ky\rVert^2 \le \lVert y\rVert \lVert \hat{v}_{k+1}\rVert\lVert V_ky\rVert$$

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