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



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