Proof that $\frac{2x}{1+x}\leq f(x) \leq \frac{1+x}{2}$ for standard operator monotone function $f$ I am reading a text on operator monotones, defined as

Definition 1 (Operator Monotone) A function $f:I\to\mathbb{R}$ defined on an interval $I \subset \mathbb{R}$ is said to be operator monotone if $$A \geq B \geq 0 \implies f(A) \geq f(B) \geq 0$$for every pair of Hermitian matrices $A$ and $B$ with spectra in $I$.

Here the $\geq$ symbols are interpreted as $A\geq B \implies A-B$  is positive semi-definite. The above definition makes sense to me, via functional calculus. The author also defines,

Definition 2 (Operator Concave) A function $f:I \to \mathbb{R}$ defined on interval $I\subset \mathbb{R}$ is said to be operator concave if $$f(\lambda A + (1-\lambda) B) \geq \lambda f(A) + (1-\lambda) f(B)$$ for every $\lambda \in [0, 1]$ and every pair of Hermitian matrices $A$ and $B$ with spectra in $I$.

and states

Theorem 1 A function $f:(0, \infty) \to (0, \infty)$ is operator monotone if and only if it is operator concave.

Finally,

Definition 3 (Standard Operator Monotone) A positive operator monotone function $f:\mathbb{R}^+ \to \mathbb{R}^+$ satisfying $$f(t) = t f\left (\frac{1}{t}\right), \hspace{1cm}  t>0$$ is called a standard operator monotone. We shall henceforth normalise $f$ so that $f(1)=1$.

Now the author states a theorem (without proof) that I am not able to prove in any way

Theorem 2 Given a standard operator monotone function $f:\mathbb{R}^+\to\mathbb{R}^+$, $f$ lies between the arithmetic and harmonic means, $$\frac{2x}{1+x} \leq f(x) \leq \frac{1+x}{2}$$

How do I go about proving this last theorem? The only idea I have is that $f$ should be a monotonic function on $\mathbb{R}$, be concave on $\mathbb{R}$ and satisfy $f(t) = t f(1/t)$ for $t>0$. Any help is much appreciated and of course, this is not a homework problem.
 A: We know that $f(1)$ is $1$ and we have a functional equation relating $f(t)$ to $f(\frac1t)$: values of $f$ on either side of $1$. So it makes sense to use the concavity of $f$ to relate $f(1)$ to $f(t)$ and $f(\frac1t)$.
We have $1 = \frac1{t+1} \cdot t + (1 - \frac1{t+1}) \cdot \frac1t$ and therefore
$$
    f(1) \ge \frac1{t+1} f(t) + \frac{t}{t+1} f(\tfrac1t).
$$
On the left-hand side, we have $f(1)=1$; on the right-hand side, we have $f(\frac1t) = \frac{f(t)}{t}$. So we can turn all of this into an upper bound on $f(t)$, which turns out to be $\frac{1+t}{2}$ when you do the algebra. This gives us the upper bound.
The lower bound $f(x) \ge \frac{2x}{1+x}$ is equivalent to the upper bound $g(x) \le \frac{1+x}{2}$ for the function $g(t) = \frac1{f(1/t)}$. According to this page, if $f$ is operator monotone, then so is $g$, but I confess that I have no idea how to prove that. We can also check that if $f$ is standard operator monotone, then so is $g$. Therefore the lower bound follows by applying our previous work to $g$.
(The second part, but not the first, really uses the "operator" part of the definitions: I don't think it would just be true for concave/monotone functions $f: (0,\infty) \to (0,\infty)$.)
