I am not an expert, bu I know that Chernoff can be optimized. Let $\mathbb{E}[X]=\mu,$ then or any positive $\Delta,$ we have $$ \mathbb{P}[X\geq E[X]+\Delta]\leq e^\Delta\left(\frac{\mu}{\mu+\Delta}\right)^{\mu+\Delta}, $$ and $$ \mathbb{P}[X\leq E[X]-\Delta]\leq e^{-\Delta}\left(\frac{\mu}{\mu\Delta}\right)^{\mu-\Delta}.\quad $$ Similarly for any positive $\delta,$ we have $$ \mathbb{P}[X\geq E[X](1+\delta)]\leq \left(\frac{e^\delta}{(1+\delta)^{1+\delta}}\right)^\mu, $$ and $$ \mathbb{P}[X\leq E[X](1-\delta)]\leq \left(\frac{e^{-\delta}}{(1-\delta)^{1-\delta}}\right)^\mu. $$ For simplicity, I will consider a standard simplification (and weakening) of the multiplicative bounds, which is a bit weaker. For any $\delta \in (0,1),$ we have $$ \mathbb{P}[X\geq E[X](1+\delta)]\leq \exp[-\delta^2 \mu/3] $$ and $$ \mathbb{P}[X\leq E[X](1-\delta)]\leq \exp[-\delta^2 \mu/2] $$ As an example, if $\mu=\mathbb{E}[X]=\sqrt{n},$ then we obtain (by taking the weaker lower tail) $$ \mathbb{P}[|X- \mu|\geq \delta \mu]\leq \exp[-\delta^2 \mu/2], $$ or equivalently letting $x=\delta \mu,$ $$ \mathbb{P}[|X- \mu|\geq x]\leq \exp[-(x^2/\mu^2) \mu/2]=\exp[-x^2/2\mu]= \exp\left[\frac{-x^2}{2\sqrt{n}}\right]. $$ So how tight the bound is depends on the exact value of $\mu$ and the value of $x,$ the distance from the expectation. If $x=c \sqrt{n},$ so you look at bounding departures of the order of the mean you get an upper bound of the form $\exp[-c \sqrt{n}].$

I am not an expert, bu I know that Chernoff can be optimized. Let $\mathbb{E}[X]=\mu,$ then or any positive $\Delta,$ we have $$ \mathbb{P}[X\geq E[X]+\Delta]\leq e^\Delta\left(\frac{\mu}{\mu+\Delta}\right)^{\mu+\Delta}, $$ and $$ \mathbb{P}[X\leq E[X]-\Delta]\leq e^{-\Delta}\left(\frac{\mu}{\mu\Delta}\right)^{\mu-\Delta}.\quad $$ Similarly for any positive $\delta,$ we have $$ \mathbb{P}[X\geq E[X](1+\delta)]\leq \left(\frac{e^\delta}{(1+\delta)^{1+\delta}}\right)^\mu, $$ and $$ \mathbb{P}[X\leq E[X](1-\delta)]\leq \left(\frac{e^{-\delta}}{(1-\delta)^{1-\delta}}\right)^\mu. $$ For simplicity, I will consider a standard simplification (and weakening) of the multiplicative bounds, which is a bit weaker. For any $\delta \in (0,1),$ we have $$ \mathbb{P}[X\geq E[X](1+\delta)]\leq \exp[-\delta^2 \mu/3] $$ and $$ \mathbb{P}[X\leq E[X](1-\delta)]\leq \exp[-\delta^2 \mu/2] $$ As an example, if $\mu=\mathbb{E}[X]=\sqrt{n},$ then we obtain (by taking the weaker lower tail) $$ \mathbb{P}[|X- \mu|\geq \delta \mu]\leq 2\exp[-\delta^2 \mu/3], $$ or equivalently letting $x=\delta \mu,$ $$ \mathbb{P}[|X- \mu|\geq x]\leq 2\exp[-(x^2/\mu^2) \mu/3]=2\exp[-x^2/3\mu]= 2\exp\left[\frac{-x^2}{3\sqrt{n}}\right]. $$ So how tight the bound is depends on the exact value of $\mu$ and the value of $x,$ the distance from the expectation. If $x=c \sqrt{n},$ so you look at bounding departures of the order of the mean you get an upper bound of the form $\exp[-c \sqrt{n}].$

Chernoff can be optimized. Let $\mathbb{E}[X]=\mu,$ then or any positive $\Delta,$ we have $$ \mathbb{P}[X\geq E[X]+\Delta]\leq e^\Delta\left(\frac{\mu}{\mu+\Delta}\right)^{\mu+\Delta}, $$ and $$ \mathbb{P}[X\leq E[X]-\Delta]\leq e^{-\Delta}\left(\frac{\mu}{\mu\Delta}\right)^{\mu-\Delta}.\quad $$ Similarly for any positive $\delta,$ we have $$ \mathbb{P}[X\geq E[X](1+\delta)]\leq \left(\frac{e^\delta}{(1+\delta)^{1+\delta}}\right)^\mu, $$ and $$ \mathbb{P}[X\leq E[X](1-\delta)]\leq \left(\frac{e^{-\delta}}{(1-\delta)^{1-\delta}}\right)^\mu. $$ For simplicity, I will consider a standard simplification (and weakening) of the multiplicative bounds, which is a bit weaker. For any $\delta \in (0,1),$ we have $$ \mathbb{P}[X\geq E[X](1+\delta)]\leq \exp[-\delta^2 \mu/3] $$ and $$ \mathbb{P}[X\leq E[X](1-\delta)]\leq \exp[-\delta^2 \mu/2] $$ So if $\mu=\mathbb{E}[X]=\sqrt{n},$ then we obtain (by taking the weaker lower tail) $$ \mathbb{P}[|X- \mu|\geq \delta \mu]\leq \exp[-\delta^2 \mu/2], $$ or equivalently letting $x=\delta \mu,$ $$ \mathbb{P}[|X- \mu|\geq x]\leq \exp[-(x^2/\mu^2) \mu/2]=\exp[-x^2/2\mu]= \exp\left[\frac{-x^2}{2\sqrt{n}}\right]. $$ So how tight the bound is depends on the exact value of $\mu$ and the value of $x,$ the distance from the expectation. If $x=c \sqrt{n},$ so you look at bounding departures of the order of the mean you get an upper bound of the form $\exp[-c \sqrt{n}].$

I am not an expert, bu I know that Chernoff can be optimized. Let $\mathbb{E}[X]=\mu,$ then or any positive $\Delta,$ we have $$ \mathbb{P}[X\geq E[X]+\Delta]\leq e^\Delta\left(\frac{\mu}{\mu+\Delta}\right)^{\mu+\Delta}, $$ and $$ \mathbb{P}[X\leq E[X]-\Delta]\leq e^{-\Delta}\left(\frac{\mu}{\mu\Delta}\right)^{\mu-\Delta}.\quad $$ Similarly for any positive $\delta,$ we have $$ \mathbb{P}[X\geq E[X](1+\delta)]\leq \left(\frac{e^\delta}{(1+\delta)^{1+\delta}}\right)^\mu, $$ and $$ \mathbb{P}[X\leq E[X](1-\delta)]\leq \left(\frac{e^{-\delta}}{(1-\delta)^{1-\delta}}\right)^\mu. $$ For simplicity, I will consider a standard simplification (and weakening) of the multiplicative bounds, which is a bit weaker. For any $\delta \in (0,1),$ we have $$ \mathbb{P}[X\geq E[X](1+\delta)]\leq \exp[-\delta^2 \mu/3] $$ and $$ \mathbb{P}[X\leq E[X](1-\delta)]\leq \exp[-\delta^2 \mu/2] $$ As an example, if $\mu=\mathbb{E}[X]=\sqrt{n},$ then we obtain (by taking the weaker lower tail) $$ \mathbb{P}[|X- \mu|\geq \delta \mu]\leq \exp[-\delta^2 \mu/2], $$ or equivalently letting $x=\delta \mu,$ $$ \mathbb{P}[|X- \mu|\geq x]\leq \exp[-(x^2/\mu^2) \mu/2]=\exp[-x^2/2\mu]= \exp\left[\frac{-x^2}{2\sqrt{n}}\right]. $$ So how tight the bound is depends on the exact value of $\mu$ and the value of $x,$ the distance from the expectation. If $x=c \sqrt{n},$ so you look at bounding departures of the order of the mean you get an upper bound of the form $\exp[-c \sqrt{n}].$