One answer - subgaussian variables generalize this property.

Let $\mu = \sum_i p_i z_i$, then the distribution is considered $\sigma^2$-subgaussian if for all $\lambda \in \mathbb{R}$,

$$ \log\left(\sum_i p_i e^{\lambda(z_i - \mu)}\right) \leq \frac{\lambda^2 \sigma^2}{2} $$

i.e.

$$ \log\left(e^{-\lambda \mu} \sum_i p_i e^{\lambda z_i}\right) \leq \frac{\lambda^2 \sigma^2}{2} $$

i.e.

$$ \log\left(\sum_i p_i e^{\lambda z_i}\right) - \lambda \mu \leq \frac{\lambda^2 \sigma^2}{2} . $$

Your expression is the case $\lambda=1$.

(In a sense this is not an answer, just a definition for when and how your inequality can be satisfied, but hopefully useful.)

One answer - subgaussian variables generalize this property.

Let $\mu = \sum_i p_i z_i$, then the distribution is considered $\sigma^2$-subgaussian if for all $\lambda \in \mathbb{R}$,

$$ \log\left(\sum_i p_i e^{\lambda(z_i - \mu)}\right) \leq \frac{\lambda^2 \sigma^2}{2} $$

i.e.

$$ \log\left(e^{-\lambda \mu} \sum_i p_i e^{\lambda z_i}\right) \leq \frac{\lambda^2 \sigma^2}{2} $$

i.e.

$$ \log\left(\sum_i p_i e^{\lambda z_i}\right) - \lambda \mu \leq \frac{\lambda^2 \sigma^2}{2} . $$

Your expression is the case $\lambda=1$.

In a sense this is not an answer, just a definition for when and how your inequality can be satisfied, but hopefully useful because we know ways to show variables are subgaussian:

1. Any random variable bounded in $[a,b]$ is $\left(\frac{b-a}{2}\right)^2$-subgaussian.
2. If $X$ is $\sigma^2$-subgaussian, then $cX$ is $c^2\sigma^2$-subgaussian.
3. If $X$ and $Y$ are independent and $\sigma_1^2,\sigma_2^2$ subgaussian, then $X+Y$ is $\sigma_1^2 + \sigma_2^2$ subgaussian.