Suppose we have a classical quantum state $\sum_x |x\rangle \langle x|\otimes \rho_x$, one can define the smooth-min entropy $H_\min(A|B)_\rho$ as the best probability of guessing outcome $x$ given $\rho_x$. How does this quantity relate to $H(A|B)_\rho$ the standard conditional entropy? If not, how does it relate to the mutual information $I(A:B)_\rho$?
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The conditional min-entropy $\text{H}_{\text{min}}(A | B)_{\rho}$ can be defined for an arbitrary state $\rho$ of a pair of registers $(A,B)$ as $$ - \inf_{\sigma} \,\text{D}_{\text{max}}(\rho \| \mathbb{1}\otimes \sigma), $$ where the infimum is over all states $\sigma$ of $B$ and $\text{D}_{\text{max}}$ is the quantum relative max-entropy: $$ \text{D}_{\text{max}}(P\|Q) = \inf\{\lambda\in\mathbb{R}: P\leq 2^{\lambda} Q\}. $$ In contrast, the ordinary conditional entropy $\text{H}(A | B)_{\rho}$ can be expressed as $$ - \inf_{\sigma}\, \text{D}(\rho \| \mathbb{1}\otimes \sigma), $$ where here $\text{D}$ refers to the ordinary quantum relative entropy. (This expression for the conditional entropy simplifies to something more familiar once you know that the infimum is always achieved by $\sigma = \operatorname{Tr}_{A}(\rho)$, which is not necessarily true for the formula for the conditional min-entropy.)
It happens to be the case that for a classical-quantum state $$ \rho = \sum_x p(x) \,|x\rangle \langle x | \otimes \rho_x $$ that the conditional min-entropy $\text{H}_{\text{min}}(A | B)_{\rho}$ is equal to the negative logarithm of the optimal guessing probability.
It is always the case that $$ \text{H}_{\text{min}}(A | B)_{\rho} \leq \text{H}(A | B)_{\rho}, $$ for every state $\rho$ and not just classical-quantum states. This follows from the fact that $$ \text{D}(\rho \| Q) \leq \text{D}_{\text{max}}(\rho \| Q) $$ for every density operator $\rho$ and every positive semidefinite operator $Q$. This inequality follows from the observation that $\rho \leq 2^{\lambda} Q$ implies \begin{align} \text{D}(\rho \| Q) & = \operatorname{Tr}(\rho \log(\rho)) - \operatorname{Tr}(\rho \log(Q))\\ & \leq \operatorname{Tr}(\rho\log(\rho)) - \operatorname{Tr}(\rho\log(2^{-\lambda}\rho))\\ & = \lambda, \end{align} where the inequality makes use of the operator monotonicity of the logarithm function: if $Q \geq 2^{-\lambda}\rho$, then $\log(Q) \geq \log(2^{-\lambda}\rho)$.
