Why $\|(A + cI)^{-1}x\|\leq \frac{\|x\|}{\lambda_{\min}(A)}$

Question

Given $A\succ 0$ (positive-definite) and $c>0$, I am trying to show

$$\|(A + cI)^{-1}x\|\leq \frac{\|x\|}{\lambda_{\min}(A)} \tag{1}$$

using information like this but without success so far. Could you please help to prove it? Any help is welcome.

EDIT: This inequality is an abstract form of the inequality

$$\left\| \left[f''(x_k) - r_M(x_k)\frac{M}{2}I \right]^{-1}f'(x_k)\right\| \leq \frac{\|f'(x_k)\|}{\lambda_{min}\left(f''(x_k)\right)}\tag{2}$$ appeared here (Theorem 3). Here is the part of the proof

Answer 1

Since $A\succ 0$, so is $A+cI\succ 0$, so $\lVert (A+cI)^{-1}\rVert = \lambda_{\min}(A + cI)^{-1}$, and $$ \lVert (A+cI)^{-1}x\rVert\leq \lVert (A+cI)^{-1}\rVert\lVert x\rVert = \frac{\lVert x\rVert}{\lambda_{\min}(A+cI)}\leq \frac{\lVert x\rVert}{\lambda_{\min}(A)}. $$

Answer 2

This is not true. E.g. when $c>0$ is small, $A=\operatorname{diag}(1,c)$ and $x=(0,1)^T$, we have $$ \|(A+cI)^{-1}x\|=\frac{1}{2c}>1=\frac{\|x\|}{\lambda_\max(A)}. $$

Answer 3

Write $$(A+cI)y=x$$ What you wish to show is $$ \| y \| \leq \frac{1}{\lambda_{min}(A)}\| x\|. $$ Let us show a little bit more, namely; $$ \| y \| \leq \frac{1}{\lambda_{min}(A)+c}\| x\|. $$ By definition of positive definiteness, $$ Ay \cdot y \geq \lambda_\min(A) \| y \|^2 $$ And of course, $A+cI_d$ is also positive definite, and $$ Ay \cdot y \geq \left(\lambda_\min(A)+c \right)\| y \|^2 $$.

By Cauchy-Schwarz $$ \| y \| \| x \| \geq x \cdot y $$ So $$ \| y \| \| x \| \geq x \cdot y = Ay \cdot y \geq \left(\lambda_\min(A)+c \right)\| y \|^2 $$ If $x=0$, then $y=0$, and the result is trivial. If not, then $y\neq0$, and we obtain the result by simplifying by $\|y\|$.