[proofplan] Apply the Jacobi identity for the Poisson bracket to the triple $(f, g, H)$. The hypotheses $\{f, H\} = 0$ and $\{g, H\} = 0$ kill two of the three terms, leaving $\{H, \{f, g\}\} = 0$, which says $\{f, g\}$ is a first integral. [/proofplan] [step:Apply the Jacobi identity to $f$, $g$, and $H$] The Jacobi identity for the Poisson bracket states that for any three smooth functions $f, g, H: M \to \mathbb{R}$, \begin{align*} \{f, \{g, H\}\} + \{g, \{H, f\}\} + \{H, \{f, g\}\} = 0. \end{align*} [/step] [step:Use the first-integral hypotheses to eliminate two terms] Since $f$ is a first integral, $\{f, H\} = 0$. By antisymmetry, $\{H, f\} = -\{f, H\} = 0$. Therefore the second term vanishes: \begin{align*} \{g, \{H, f\}\} = \{g, 0\} = 0. \end{align*} Since $g$ is a first integral, $\{g, H\} = 0$, so the first term also vanishes: \begin{align*} \{f, \{g, H\}\} = \{f, 0\} = 0. \end{align*} The Jacobi identity now reduces to \begin{align*} 0 + 0 + \{H, \{f, g\}\} = 0, \end{align*} i.e., $\{H, \{f, g\}\} = 0$. By antisymmetry, $\{\{f, g\}, H\} = -\{H, \{f, g\}\} = 0$, confirming that $\{f, g\}$ is a first integral. [/step]