I just proved an elementary inequality that
\begin{align*}
\log x\geq\dfrac{2(x-1)}{x+1},~~~~x\geq 1.
\end{align*}
Writing in the integral form we see that
\begin{align*}
\dfrac{1}{x-1}\int_{1}^{x}\dfrac{1}{t}dt\geq\dfrac{1}{\dfrac{x+1}{2}},~~~~x>1.
\end{align*}
Somehow I just have some mind that whether
\begin{align*}
\dfrac{1}{x-1}\int_{1}^{x}\dfrac{1}{\varphi(t)}dt\geq\dfrac{1}{\varphi\left(\dfrac{x+1}{2}\right)},~~~~x>1
\end{align*}
is true for general nonnegative, increasing, continuous function $\varphi$ on $[1,\infty)$.
I have tried $\varphi(t)=t^{2}$, the integral inequality still works.
Assume further that $\varphi$ is differentiable, I tried to prove the inequality using the standard trick that
\begin{align*}
\eta(x)=\dfrac{1}{x-1}\int_{1}^{x}\dfrac{1}{\varphi(t)}dt-\dfrac{1}{\varphi\left(\dfrac{x+1}{2}\right)},~~~~x>1
\end{align*}
and letting that $\eta(1)=0$ to show that $\eta'(x)\geq 0$, there is nothing I can conclude.
If the integral inequality is true, what about some abstract higher dimensional generalization?
\begin{align*}
\dfrac{1}{\mu(S)}\int_{S}\dfrac{1}{\varphi(t)}dt\geq\dfrac{1}{\varphi(m)},
\end{align*}
where $m$ is some sort of mean value, so to speak, in the case of the integral inequality, we choose
\begin{align*}
m=\dfrac{1}{x-1}\int_{1}^{x}tdt=\dfrac{x+1}{2},
\end{align*}
where in one-dimensional, perhaps we choose
\begin{align*}
m=\dfrac{1}{\mu(S)}\int_{S}tdt.
\end{align*}
I have no idea what is the best choice of such a mean value in higher dimensional.
Best Answer
A partial answer: in the special case that the map $t\mapsto 1/\varphi(t)$ is convex on the interval $(1,\infty)$, then this holds for all $x>1$. This is just Jensen's Inequality applied to $U((1,x))$, i.e. a uniformly distributed random variable on $(1,x)$. Indeed, your equation then just reads $\mathbb{E}[1/\varphi(X)]\geq 1/\varphi(\mathbb{E}[X])$. This is the case when $\phi(t)=t^k$ for any $k\geq 1$.
To find a sufficient condition on just $\varphi$, we have \begin{equation} \frac{d^2}{dt^2} \frac{1}{\varphi(t)}=\frac{2\varphi'(t)^2-\varphi(t)\varphi''(t)}{\phi(t)^3}, \end{equation} so a sufficient condition is just that $2\phi'(t)\geq \varphi(t)\varphi''(t)$ for all $t\in (0,\infty)$, as this ensures convexity because the denominator is assumed to be positive. Of course, this assumed twice differentiability, but the moral is that it is sufficient that $\varphi$ be concave, as $\varphi$ is increasing. This characterization should extend even when nondifferentiable. It generalizes to higher dimensions where $m$ is taken to be the expected value of a random variable uniformly distributed in your region.
I'll think about counterexamples when this doesn't hold. EDIT: here's an example where this fails: take $\phi(t)=1+(t-1)^2$. This satisfies your conditions. But one can check that for $t\in [1,1.5]$, the function $1/\phi(t)$ is concave, so the reverse inequality necessarily holds by Jensen's Inequality. Explicitly, we have \begin{equation} \frac{1}{1.5-1} \int_1^{1.5} \frac{1}{1+(t-1)^2}dt\approx .927 < .9411 \approx \frac{1}{1+((1.5+1)/2 -1)^2}=\frac{1}{1+.25^2}. \end{equation}