[proofplan] The proof is a deterministic comparison between Bernstein's quadratic-linear exponent and the minimum of its quadratic and linear regimes. We prove the exponent comparison \begin{align*} \frac{t^2}{2\left(V+\frac{bt}{3}\right)} \ge \frac{3}{8}\min\left\{\frac{t^2}{V},\frac{t}{b}\right\} \end{align*} by splitting according to whether $V$ is at least $bt$ or smaller than $bt$. Substituting this lower bound into the assumed Bernstein bound gives the stated minimum form. [/proofplan] [step:Handle the endpoint $t=0$] For $t=0$, the event $\{\sum_{i=1}^n X_i \ge 0\}$ has probability at most $1$. By the convention in the statement, \begin{align*} \exp\left( -\frac{3}{8}\min\left\{\frac{0^2}{V},\frac{0}{b}\right\} \right)=\exp(0)=1. \end{align*} Thus the desired inequality holds for $t=0$. [/step] [step:Compare Bernstein's exponent with the quadratic regime when $V\ge bt$] Fix $t>0$. Suppose first that $V\ge bt$. Since $b>0$ and $t>0$, this implies $V>0$, so $\frac{t^2}{V}$ is finite. From $bt\le V$ we obtain \begin{align*} 2\left(V+\frac{bt}{3}\right) \le 2\left(V+\frac{V}{3}\right) = \frac{8V}{3}. \end{align*} Taking reciprocals of positive quantities and multiplying by $t^2$ gives \begin{align*} \frac{t^2}{2\left(V+\frac{bt}{3}\right)} \ge \frac{3t^2}{8V} = \frac{3}{8}\frac{t^2}{V} \ge \frac{3}{8}\min\left\{\frac{t^2}{V},\frac{t}{b}\right\}. \end{align*} [guided] Fix $t>0$ and assume $V\ge bt$. This is the regime in which the variance term controls the denominator in Bernstein's exponent. Since $b>0$ and $t>0$, the inequality $V\ge bt$ gives $V>0$, so the expression $\frac{t^2}{V}$ is a genuine finite real number. The point is to replace the mixed denominator $V+\frac{bt}{3}$ by a constant multiple of $V$. The assumption $bt\le V$ gives \begin{align*} V+\frac{bt}{3} \le V+\frac{V}{3} = \frac{4V}{3}. \end{align*} Multiplying by $2$ yields \begin{align*} 2\left(V+\frac{bt}{3}\right) \le \frac{8V}{3}. \end{align*} All quantities are positive, so taking reciprocals reverses the inequality. Multiplying by $t^2>0$ gives \begin{align*} \frac{t^2}{2\left(V+\frac{bt}{3}\right)} \ge \frac{3t^2}{8V} = \frac{3}{8}\frac{t^2}{V}. \end{align*} Finally, the minimum of two numbers is at most each of them, so \begin{align*} \frac{3}{8}\frac{t^2}{V} \ge \frac{3}{8}\min\left\{\frac{t^2}{V},\frac{t}{b}\right\}. \end{align*} Combining the displayed inequalities proves the required lower bound for Bernstein's exponent in the quadratic regime. [/guided] [/step] [step:Compare Bernstein's exponent with the linear regime when $V<bt$] Now suppose $V<bt$. Under the Bernstein hypotheses, the variance proxy is \begin{align*} V=\sum_{i=1}^n \mathbb E[X_i^2]. \end{align*} Each [random variable](/page/Random%20Variable) $X_i^2$ is non-negative, so each expectation $\mathbb E[X_i^2]$ is non-negative, and therefore $V\ge 0$. Since $V\ge 0$, $b>0$, and $t>0$, we have \begin{align*} 2\left(V+\frac{bt}{3}\right) < 2\left(bt+\frac{bt}{3}\right) = \frac{8bt}{3}. \end{align*} Taking reciprocals of positive quantities and multiplying by $t^2$ gives \begin{align*} \frac{t^2}{2\left(V+\frac{bt}{3}\right)} > \frac{3t^2}{8bt} = \frac{3}{8}\frac{t}{b} \ge \frac{3}{8}\min\left\{\frac{t^2}{V},\frac{t}{b}\right\}, \end{align*} where, if $V=0$, the convention $\frac{t^2}{V}=\infty$ is used. [/step] [step:Substitute the exponent comparison into the assumed Bernstein bound] Combining the two cases, for every $t>0$ we have \begin{align*} \frac{t^2}{2\left(V+\frac{bt}{3}\right)} \ge \frac{3}{8}\min\left\{\frac{t^2}{V},\frac{t}{b}\right\}. \end{align*} The assumed Bernstein bound gives, for every $t>0$, \begin{align*} \mathbb P\left(\sum_{i=1}^n X_i \ge t\right) \le \exp\left(-\frac{t^2}{2\left(V+\frac{bt}{3}\right)}\right). \end{align*} The exponential map $s\mapsto e^{-s}$ is decreasing on $[0,\infty)$, so the exponent comparison gives \begin{align*} \exp\left(-\frac{t^2}{2\left(V+\frac{bt}{3}\right)}\right) \le \exp\left( -\frac{3}{8}\min\left\{\frac{t^2}{V},\frac{t}{b}\right\} \right). \end{align*} Combining the last two displayed inequalities gives the desired estimate for every $t>0$. Together with the endpoint $t=0$, this proves the statement with the universal constant \begin{align*} c=\frac{3}{8}. \end{align*} [/step]