Solving a probability inequality Given three probability density functions $p$, $q$ and $h$ of Normal distributions. $p$ and $h$ are fixed and I want to solve for $q$, is it possible to identify a condition on $q$ apart from the trivial cases q=h or q=p such that:
$$\int_{z} \frac{p(z)}{q(z)} \: h(z) \: dz \leqslant 1$$

Another approach: WLOG assume $p \sim N(a, \sigma_1^2)$ and $h \sim N(b, \sigma_2^2)$ where $0<\sigma_1^2 \leq \sigma_2^2$. Define $q\sim N(r, \sigma_2^2)$ for some $r \in \mathbb{R}$ that we will use to achieve the result.
Then
\begin{align}
\frac{h(z)}{q(z)} &= \exp\left(-\frac{(z-b)^2}{2\sigma_2^2}\right)\exp\left(\frac{(z-r)^2}{2\sigma_2^2}\right) \\
&= \exp\left(-\frac{1}{2\sigma_2^2}\left[(z-b)^2 - (z-r)^2 \right]\right)\\
&=\exp\left(\frac{1}{2\sigma_2^2}\left[-2z(b-r) + b^2 -r^2 \right]\right)
\end{align}
So then if we define $\lambda = \sigma_1^2/\sigma_2^2$, we have $0<\lambda \leq 1$ and
\begin{align}
\frac{p(z)h(z)}{q(z)}&=\frac{1}{\sqrt{2\pi\sigma_1^2}}\exp\left(-\frac{(z-a)^2}{2\sigma_1^2}\right)\exp\left(\frac{1}{2\sigma_2^2}\left[-2z(b-r) + b^2 -r^2 \right]\right)\\
&=\frac{1}{\sqrt{2\pi \sigma_1^2}}\exp\left(-\frac{1}{2\sigma_1^2}\left[ (z-a)^2 + 2\lambda z(b-r) -\lambda b^2 + \lambda r^2\right]\right)\\
&=\frac{1}{\sqrt{2\pi \sigma_1^2}}\exp\left(-\frac{1}{2\sigma_1^2}\left[ (z-a+\lambda (b-r))^2 - (-a+\lambda(b-r))^2 - \lambda b^2 + \lambda r^2\right]\right)\\
\end{align}
So
\begin{align}
\int_{-\infty}^{\infty} \frac{p(z)h(z)}{q(z)}dz &= \exp\left(\frac{1}{2\sigma_1^2}\left[(-a+\lambda(b-r))^2 +\lambda b^2- \lambda r^2\right]\right)\\
&=\exp\left(\frac{1}{2\sigma_1^2}\left[ (\lambda^2-\lambda)r^2 + 2ar\lambda + d\right]\right)\\
\end{align}
for some $d \in \mathbb{R}$.  We know $0\leq \lambda \leq 1$. If $0<\lambda<1$ then $\lambda^2-\lambda < 0$ and we can choose $r$ suitably large to make the result strictly less than 1.  If $\lambda=1$ then, if $a\neq 0$, we can again choose $r$ to make the result strictly less than 1.
The only remaining case is if $\lambda=1$ and $a=0$, which gives
\begin{align}
\int_{-\infty}^{\infty} \frac{p(z)h(z)}{q(z)}dz &= \exp\left(\frac{1}{2\sigma_1^2}\left[(-a+\lambda(b-r))^2 +\lambda b^2- \lambda r^2\right]\right)\\
&=\exp\left(\frac{1}{2\sigma_1^2}\left[(b-r)^2+b^2-r^2\right]\right)\\
&=\exp\left(\frac{1}{2\sigma_1^2}\left[2b(b-r)\right]\right)\\
\end{align}
and if $b\neq 0$ we can choose $r$ to make the result strictly less than 1.
The only remaining remaining case is when $\lambda=1$ and $a=b=0$, meaning that we have $p =h\sim N(0,\sigma^2)$.
