Consider a symmetric positive definite matrix $R > 0 $ and a symmetric positive semi-definite matrix $P \ge 0$. Does this hold?
$$\|(R + P)^{-1} \| \le \| R^{-1} \|? $$
and does this hold too?
$$\|(R + P)^{-1} \| \le \| R^{-1} \| \le \frac{1}{\| R \|}? $$
Yes, it holds. It is a non-trivial fact that if $A,B$ are real and symmetric (more generally, selfadjoint) and $A\leq B$ (i.e. $B-A$ is positive semi-definite) then $\lambda_j(A)\leq\lambda_j(B)$, where $\lambda_1(A)\geq\lambda_2(A)\geq\cdots\geq\lambda_n(A)$ are the eigenavalues (counting multiplicities).
From the above, then $\lambda_n(R+P)\geq\lambda_n(R)$, so (with the operator norm) $$ \|R^{-1}\|=\frac1{\lambda_n(R)}\geq\frac1{\lambda_n(R+P)}=\|(R+P)^{-1}\|. $$
More generally, for any Ky-Fan norm, $$ \|A\|_{(k)}=\sum_{j=1}^k\lambda_j(A)^\downarrow, $$ since we are dealing with positive matrices the eigenvalues are the singular values. So $$ \|(R+P)^{-1}\|_{(k)}=\sum_{j=n-k+1}^n\frac1{\lambda_j(R+P)}\leq \sum_{j=n-k+1}^n\frac1{\lambda_j(R)}=\|R^{-1}\|_{(k)}. $$ As the inequality holds for all Ky-Fan norms, it holds for all unitarily invariant norms.
The inequality $$\tag{*}\|R^{-1}\|\leq\frac1{\|R\|}$$ can only hold for the spectral norm if $\|R\|\,\|R^{-1}\|=1$ (because you always have $\|R\|\,\|R^{-1}\|\geq1$). This equality is $\lambda_1(R)/\lambda_n(R)=1$, i.e. $\lambda_1(R)=\lambda_n(R)$. So the inequality $(*)$ holds only for scalar multiplies of the identity (and, in such case, it is an equality).