Since the OP was kind enough to reference an older answer of mine, and also to alert me to the fact, I will provide some input here, of naive flavor, I guess, since mathoverflow is definitely over and out of my league. I will use $\Phi()$ for the standard normal CDF and $\phi()$ for the standard normal PDF. In my answer in math.se I had proved that

$${\rm Var}(Y) = {\rm Var}(X)\cdot \left[1+\sigma^2\frac{\partial^2 \ln H(\mu)}{\partial \mu^2}\right],\;\; -1 <\sigma^2\frac{\partial^2 \ln H(\mu)}{\partial \mu^2} \leq 0 $$

where:

$$\ln H(\mu)=\ln \big(\Phi(\beta(\mu))-\Phi(\alpha(\mu))\big)$$

$$\alpha=(a-\mu)/\sigma, \;\beta=(b-\mu)/\sigma$$.

The weak inequality comes from the fact that $H$ is log-concave (see the original post).

Calculating the second derivative, in order to prove strict inequality, we have to show that

$$\frac{\partial^2 \ln H(\mu)}{\partial \mu^2} < 0 \implies D \equiv [-\phi'(\alpha)+\phi'(\beta)][\Phi(\beta)-\Phi(\alpha)]-[\phi(\alpha)-\phi(\beta)]^2 < 0 $$

We note that $\Phi(\beta)-\Phi(\alpha) >0 $ always.

CASE A : $a \leq \mu \leq b$ (including one-sided trunctaions).

Here, by the unimodality of $\phi()$, with $\mu$ being the mode, we have that

$$\phi'(\alpha) \geq 0, \phi'(\beta) \leq 0 \implies -\phi'(\alpha)+\phi'(\beta) < 0$$

since the two derivatives cannot be both equal to zero.

Then $D < 0$ always (including truncation symmetric around the mean, where, due to the fact that $\phi()$ is an even function, we have that $\phi(\alpha)=\phi(\beta)$, and the second element of $D$ will be zero. But the first element is always strictly negative).
So for this case we have proved that $\partial^2 \ln H(\mu)/\partial \mu^2 < 0$ as we wanted.

The cases left out are all the cases where $\mu$ does not belong to the (two-sided or one-sided) truncated support.

CASE B : Truncated support $S_B\equiv (-\infty , b], \mu \notin S_B$

Here the inequality to prove is

$$D_B \equiv \phi'(\beta)\Phi(\beta)-\phi(\beta)^2 < 0$$

Since $\phi'(\beta) = -\beta \phi(\beta)$ we wan to show

$$-\beta \phi(\beta)\Phi(\beta)-\phi(\beta)^2 < 0 \implies -\beta \Phi(\beta)-\phi(\beta) < 0 \implies \beta \Phi(\beta)+\phi(\beta) > 0$$

But this holds because

$$\beta \Phi(\beta)+\phi(\beta) = \int_{-\infty}^{\beta}\Phi(t){\rm d}t$$

and the integral is necessarily strictly greater than zero since $\Phi()$ is non-negative and non-constantly zero. So we have proved here too what we needed to prove.

CASE C : Truncated support $S_C\equiv [a, \infty ), \mu \notin S_C$

Here the inequality to prove is

$$-\phi'(\alpha)[1-\Phi(\alpha)]-\phi(\alpha)^2 < 0 \implies \alpha \phi(\alpha)[1-\Phi(\alpha)]-\phi(\alpha)^2 < 0 $$

and simplifying and using the symmetry properties of the two functions we have

$$ \implies (-\alpha)\Phi(-\alpha) + \phi(-\alpha) >0$$ and we are at the same situation as in Case B. So QED here too.

I am not treating the two-sided truncation cases $a < b < \mu$ and $\mu < a < b$. As with the cases I treated, I am certain there exists some more advanced and elegant way to prove what we want.

Since the OP was kind enough to reference an older answer of mine, and also to alert me to the fact, I will provide some input here, of naive flavor, I guess, since mathoverflow is definitely over and out of my league. I will use $\Phi()$ for the standard normal CDF and $\phi()$ for the standard normal PDF. In my answer in math.se I had proved that

$${\rm Var}(Y) = {\rm Var}(X)\cdot \left[1+\sigma^2\frac{\partial^2 \ln H(\mu)}{\partial \mu^2}\right],\;\; -1 <\sigma^2\frac{\partial^2 \ln H(\mu)}{\partial \mu^2} \leq 0 $$

where:

$$\ln H(\mu)=\ln \big(\Phi(\beta(\mu))-\Phi(\alpha(\mu))\big)$$

$$\alpha=(a-\mu)/\sigma, \;\beta=(b-\mu)/\sigma$$.

The weak inequality comes from the fact that $H$ is log-concave (see the original post).

Calculating the second derivative, in order to prove strict inequality, we have to show that

$$\frac{\partial^2 \ln H(\mu)}{\partial \mu^2} < 0 \implies D \equiv [-\phi'(\alpha)+\phi'(\beta)][\Phi(\beta)-\Phi(\alpha)]-[\phi(\alpha)-\phi(\beta)]^2 < 0 $$

We note that $\Phi(\beta)-\Phi(\alpha) >0 $ always.

CASE A : $a \leq \mu \leq b$ (including one-sided trunctaions).

Here, by the unimodality of $\phi()$, with $\mu$ being the mode, we have that

$$\phi'(\alpha) \geq 0, \phi'(\beta) \leq 0 \implies -\phi'(\alpha)+\phi'(\beta) < 0$$

since the two derivatives cannot be both equal to zero.

Then $D < 0$ always (including truncation symmetric around the mean, where, due to the fact that $\phi()$ is an even function, we have that $\phi(\alpha)=\phi(\beta)$, and the second element of $D$ will be zero. But the first element is always strictly negative).
So for this case we have proved that $\partial^2 \ln H(\mu)/\partial \mu^2 < 0$ as we wanted.

The cases left out are all the cases where $\mu$ does not belong to the (two-sided or one-sided) truncated support.

CASE B : Truncated support $S_B\equiv (-\infty , b], \mu \notin S_B$

Here the inequality to prove is

$$D_B \equiv \phi'(\beta)\Phi(\beta)-\phi(\beta)^2 < 0$$

Since $\phi'(\beta) = -\beta \phi(\beta)$ we wan to show

$$-\beta \phi(\beta)\Phi(\beta)-\phi(\beta)^2 < 0 \implies -\beta \Phi(\beta)-\phi(\beta) < 0 \implies \beta \Phi(\beta)+\phi(\beta) > 0$$

But this holds because

$$\beta \Phi(\beta)+\phi(\beta) = \int_{-\infty}^{\beta}\Phi(t){\rm d}t$$

and the integral is necessarily strictly greater than zero since $\Phi()$ is non-negative and non-constantly zero. So we have proved here too what we needed to prove.

CASE C : Truncated support $S_C\equiv [a, \infty ), \mu \notin S_C$

Here the inequality to prove is

$$-\phi'(\alpha)[1-\Phi(\alpha)]-\phi(\alpha)^2 < 0 \implies \alpha \phi(\alpha)[1-\Phi(\alpha)]-\phi(\alpha)^2 < 0 $$

and simplifying and using the symmetry properties of the two functions we have

$$ \implies (-\alpha)\Phi(-\alpha) + \phi(-\alpha) >0$$ and we are at the same situation as in Case B. So QED here too.

I am not treating the two-sided truncation cases $a < b < \mu$ and $\mu < a < b$. As with the cases I treated, I am certain there exists some more advanced and elegant way to prove what we want.

Since the OP was kind enough to reference an older answer of mine, and also to alert me to the fact, I will provide some input here, of naive flavor, I guess, since mathoverflow is definitely over and out of my league. I will use $\Phi()$ for the standard normal CDF and $\phi()$ for the standard normal PDF. In my answer in math.se I had proved that

$${\rm Var}(Y) = {\rm Var}(X)\cdot \left[1+\sigma^2\frac{\partial^2 \ln H(\mu)}{\partial \mu^2}\right],\;\; -1 <\sigma^2\frac{\partial^2 \ln H(\mu)}{\partial \mu^2} \leq 0 $$

where:

$$\ln H(\mu)=\ln \big(\Phi(\beta(\mu))-\Phi(\alpha(\mu))\big)$$

$$\alpha=(a-\mu)/\sigma, \;\beta=(b-\mu)/\sigma$$.

The weak inequality comes from the fact that $H$ is log-concave (see the original post).

Calculating the second derivative, in order to prove strict inequality, we have to show that

$$\frac{\partial^2 \ln H(\mu)}{\partial \mu^2} < 0 \implies D \equiv [-\phi'(\alpha)+\phi'(\beta)][\Phi(\beta)-\Phi(\alpha)]-[\phi(\alpha)-\phi(\beta)]^2 < 0 $$

We note that $\Phi(\beta)-\Phi(\alpha) >0 $ always.

CASE A : $a \leq \mu \leq b$ (including one-sided trunctaions).

Here, by the unimodality of $\phi()$, with $\mu$ being the mode, we have that

$$\phi'(\alpha) \geq 0, \phi'(\beta) \leq 0 \implies -\phi'(\alpha)+\phi'(\beta) < 0$$

since the two derivatives cannot be both equal to zero.

Then $D < 0$ always (including truncation symmetric around the mean, where, due to the fact that $\phi()$ is an even function, we have that $\phi(\alpha)=\phi(\beta)$, and the second element of $D$ will be zero. But the first element is always strictly negative).
So for this case we have proved that $\partial^2 \ln H(\mu)/\partial \mu^2 < 0$ as we wanted.

The cases left out are all the cases where $\mu$ does not belong to the (two-sided or one-sided) truncated support.

CASE B : Truncated support $S_B\equiv (-\infty , b], \mu \notin S$

Here the inequality to prove is

$$D_B \equiv \phi'(\beta)\Phi(\beta)-\phi(\beta)^2 < 0$$

Since $\phi'(\beta) = -\beta \phi(\beta)$ we wan to show

$$-\beta \phi(\beta)\Phi(\beta)-\phi(\beta)^2 < 0 \implies -\beta \Phi(\beta)-\phi(\beta) < 0 \implies \beta \Phi(\beta)+\phi(\beta) > 0$$

But this holds because

$$\beta \Phi(\beta)+\phi(\beta) = \int_{-\infty}^{\beta}\Phi(t){\rm d}t$$

and the integral is necessarily strictly greater than zero since $\Phi()$ is non-negative and non-constantly zero. So we have proved here too what we needed to prove.

CASE C : Truncated support $S_C\equiv [a, \infty ), \mu \notin S_C$

Here the inequality to prove is

$$-\phi'(\alpha)[1-\Phi(\alpha)]-\phi(\alpha)^2 < 0 \implies \alpha \phi(\alpha)[1-\Phi(\alpha)]-\phi(\alpha)^2 < 0 $$

and simplifying and using the symmetry properties of the two functions we have

$$ \implies (-\alpha)\Phi(-\alpha) + \phi(-\alpha) >0$$ and we are at the same situation as in Case B. So QED here too.

I am not treating the two-sided truncation cases $a < b < \mu$ and $\mu < a < b$. As with the cases I treated, I am certain there exists some more advanced and elegant way to prove what we want.

Since the OP was kind enough to reference an older answer of mine, and also to alert me to the fact, I will provide some input here, of naive flavor, I guess, since mathoverflow is definitely over and out of my league. I will use $\Phi()$ for the standard normal CDF and $\phi()$ for the standard normal PDF. In my answer in math.se I had proved that

$${\rm Var}(Y) = {\rm Var}(X)\cdot \left[1+\sigma^2\frac{\partial^2 \ln H(\mu)}{\partial \mu^2}\right],\;\; -1 <\sigma^2\frac{\partial^2 \ln H(\mu)}{\partial \mu^2} \leq 0 $$

where:

$$\ln H(\mu)=\ln \big(\Phi(\beta(\mu))-\Phi(\alpha(\mu))\big)$$

$$\alpha=(a-\mu)/\sigma, \;\beta=(b-\mu)/\sigma$$.

The weak inequality comes from the fact that $H$ is log-concave (see the original post).

Calculating the second derivative, in order to prove strict inequality, we have to show that

$$\frac{\partial^2 \ln H(\mu)}{\partial \mu^2} < 0 \implies D \equiv [-\phi'(\alpha)+\phi'(\beta)][\Phi(\beta)-\Phi(\alpha)]-[\phi(\alpha)-\phi(\beta)]^2 < 0 $$

We note that $\Phi(\beta)-\Phi(\alpha) >0 $ always.

CASE A : $a \leq \mu \leq b$ (including one-sided trunctaions).

Here, by the unimodality of $\phi()$, with $\mu$ being the mode, we have that

$$\phi'(\alpha) \geq 0, \phi'(\beta) \leq 0 \implies -\phi'(\alpha)+\phi'(\beta) < 0$$

since the two derivatives cannot be both equal to zero.

Then $D < 0$ always (including truncation symmetric around the mean, where, due to the fact that $\phi()$ is an even function, we have that $\phi(\alpha)=\phi(\beta)$, and the second element of $D$ will be zero. But the first element is always strictly negative).
So for this case we have proved that $\partial^2 \ln H(\mu)/\partial \mu^2 < 0$ as we wanted.

The cases left out are all the cases where $\mu$ does not belong to the (two-sided or one-sided) truncated support.

CASE B : Truncated support $S_B\equiv (-\infty , b], \mu \notin S_B$

Here the inequality to prove is

$$D_B \equiv \phi'(\beta)\Phi(\beta)-\phi(\beta)^2 < 0$$

Since $\phi'(\beta) = -\beta \phi(\beta)$ we wan to show

$$-\beta \phi(\beta)\Phi(\beta)-\phi(\beta)^2 < 0 \implies -\beta \Phi(\beta)-\phi(\beta) < 0 \implies \beta \Phi(\beta)+\phi(\beta) > 0$$

But this holds because

$$\beta \Phi(\beta)+\phi(\beta) = \int_{-\infty}^{\beta}\Phi(t){\rm d}t$$

and the integral is necessarily strictly greater than zero since $\Phi()$ is non-negative and non-constantly zero. So we have proved here too what we needed to prove.

CASE C : Truncated support $S_C\equiv [a, \infty ), \mu \notin S_C$

Here the inequality to prove is

$$-\phi'(\alpha)[1-\Phi(\alpha)]-\phi(\alpha)^2 < 0 \implies \alpha \phi(\alpha)[1-\Phi(\alpha)]-\phi(\alpha)^2 < 0 $$

and simplifying and using the symmetry properties of the two functions we have

$$ \implies (-\alpha)\Phi(-\alpha) + \phi(-\alpha) >0$$ and we are at the same situation as in Case B. So QED here too.

I am not treating the two-sided truncation cases $a < b < \mu$ and $\mu < a < b$. As with the cases I treated, I am certain there exists some more advanced and elegant way to prove what we want.