Let $n \in \mathbb N$ and let $f_n$ be the function $t \mapsto t \left (t + \frac {1} {n} \right )^{-1}$ on $\mathbb R_{+}.$ Let $a$ and $b$ be two positive elements in a $C^{\ast}$-algebra $A$ such that $a \leq b$ and let $m \leq n.$ Then $f_m (a) \leq f_n (b),$ where $f_m(a)$ and $f_n(b)$ are the images of $f_m$ and $f_n$ under the continuous functional calculus for $a$ and $b$ respectively.
My Attempt $:$ Here $$f_m(a) = a \left (a + \frac {1} {m} \right )^{-1} = \left (a + \frac {1} {m} \right )^{-1} a\ \ \text{and}\ \ f_n (b) = b \left (b + \frac {1} {n} \right )^{-1} = \left (b + \frac {1} {n} \right )^{-1} b.$$
From here I couldn't quite able to show the required inequality.
Since $m \leq n$ we have $0 \leq a + \frac {1} {n} \leq a + \frac {1} {m}.$ Hence $0 \leq \left (a + \frac {1} {m} \right )^{-1} \leq \left (a + \frac {1} {n} \right )^{-1}.$ So $$\begin{align*} a \left (a + \frac {1} {m} \right )^{-1} & = a^{\frac {1} {2}} \left (a + \frac {1} {m} \right )^{-1} a^{\frac {1} {2}} \\ & \leq a^{\frac {1} {2}} \left (a + \frac {1} {n} \right )^{-1} a^{\frac {1} {2}} \\ & = a \left (a + \frac {1} {n} \right )^{-1} \\ & = \left (a + \frac {1} {n} \right )^{-\frac {1} {2}} a \left (a + \frac {1} {n} \right )^{-\frac {1} {2}} \\ & \leq \left (a + \frac {1} {n} \right )^{-\frac {1} {2}} b \left (a + \frac {1} {n} \right )^{-\frac {1} {2}} \\ & = b \left (a + \frac {1} {n} \right )^{-1} \\ & = b^{\frac {1} {2}} \left (a + \frac {1} {n} \right )^{-1} b^{\frac {1} {2}} \end{align*}$$ At this stage I got stuck. I know that $0 \leq a + \frac {1} {n} \leq b + \frac {1} {n}.$ So we have $0 \leq \left (b + \frac {1} {n} \right )^{-1} \leq \left (a + \frac {1} {n} \right )^{-1},$ which is not what I wanted to have. Could anyone give me some suggestion regarding this?
Thanks for your time.
EDIT $:$ I think the last two equalities doesn't hold in general unless $a$ and $b$ commute.
As you showed, what you want is to prove that $f_n(a)\leq f_n(b)$ if $a\leq b$. The terminology for this is that $f_n$ is operator monotone.
For notational simplicity we may take $f(t)=t(t+c)^{-1}$, with $c>0$. As a motivation, how do we show that $f$ is (number) monotone? We want to show that $$ s<t\implies\frac s{s+c}<\frac t{t+c}. $$ This is trivial if one notices that $$ \frac t{t+c}=\frac1{1+\frac ct}. $$ the big advantage being that there is a single $t$ on the right-hand-side.
To see that $f$ is operator monotone, assume first that $a$ is invertible; then so is $b$ and we have $$ f(a)=a(a+c)^{-1}=(1+ca^{-1})^{-1}. $$ Then $$ a\leq b\implies b^{-1}\leq a^{-1}\implies 1+cb^{-1}\leq 1+ca^{-1}\implies (1+ca^{-1})^{-1}\leq (1+cb^{-1})^{-1}. $$ When $a$ is not invertible, we can apply the above to $a+\varepsilon 1\leq b+\varepsilon 1$. So $$f(a)=\lim_{\varepsilon\to0}f(a+\varepsilon 1)\leq \lim_{\varepsilon\to0} f(b+\varepsilon 1)=f(b), $$ where the equalities are given by the continuity of $f$ at $t=0$. Explicitly, if $\Gamma$ is the Gelfand transform for $a$, let $g_\varepsilon(t)=f(t+\varepsilon)-f(t)$. Then $g_\varepsilon\to0$ uniformly on $\sigma(a)$, and so $$ f(a+\varepsilon)-f(a)=\Gamma(g_\varepsilon)\to0. $$
Edit: the key technical point in the argument is the fact that $0\leq a\leq b$ with $a$ invertible, implies that $b^{-1}\leq a^{-1}$. The usual proof comes from recognizing that, for $x$ positive, $x\leq 1$ if and only if $\sigma(x)\subset[0,1]$, together with the fact that $\{0\}\cup\sigma(xy)=\{0\}\cup\sigma(yx)$.
So, if $a\leq b$, then multiplying left and right by $b^{-1/2}$ we get $b^{-1/2}ab^{-1/2}\leq 1$. This is the same as $\sigma(b^{-1/2}ab^{-1/2})\subset[0,1]$. Now $$ \sigma(b^{-1/2}ab^{-1/2})=\sigma(b^{-1/2}a^{1/2}a^{1/2}b^{-1/2}) =\sigma(a^{1/2}b^{-1/2}b^{-1/2}a^{1/2})=\sigma(a^{1/2}b^{-1}a^{1/2}). $$ Thus $a^{1/2}b^{-1}a^{1/2}\leq1$. Multiplying left and right by $a^{-1/2}$ we obtain $b^{-1}\leq a^{-1}$.