$\newcommand{\si}{\sigma}\newcommand{\R}{\mathbb{R}}\newcommand{\ch}{\operatorname{ch}}$Let $X$ be a $\sigma$-sub-Gaussian random variable (r.v.) for some real $\sigma>0$, that is,
\begin{equation*}
M_X(t):=Ee^{tX}\le e^{\sigma^2t^2/2}\quad\text{for all real $t$}. \tag{1}
\end{equation*}
Then the standard upper bound on $P(X\ge u)$ is $e^{-u^2/(2\si^2)}$,
for each real $u\ge0$.

Note that (1) implies that the left and right derivatives of the moment generating function $M_X$ are $0$, and hence $EX=0$. So, the standard upper bound $e^{-u^2/(2\si^2)}$ on $P(X\ge u)$ can be indeed improved -- see Section 3.2.

However, any such improvement must be asymptotically negligible for large $u$. Specifically, for any upper bound $B(u)$ on $P(X\ge u)$ valid for all $\si$-sub-Gaussian random r.v.'s $X$ we have $B(u)\ge e^{-u^2/(2\si^2)}(1-o(1))$; here and in what follows, all asymptotic relations are given for $u\to\infty$.

Indeed, by simple rescaling, without loss of generality (wlog) $\si=1$. So, we have to show that
\begin{equation*}
P_u:=\sup_{X\in S}P(X\ge u)\ge e^{-u^2/2}(1-o(1)), \tag{2}
\end{equation*}
where $S$ is the set of all $1$-sub-Gaussian random r.v.'s.

Here is a proof of (2). For $p\in(0,1)$ and real $u>0$, let $X_{p,u}$ be any r.v. such that
\begin{equation*}
P(X_{p,u}=u)=P(X_{p,u}=-u)=p/2,\quad P(X_{p,u}=0)=1-p. \tag{2.5}
\end{equation*}
We will have $X_{p,u}\in S$ iff $1-p+p\ch tu\le e^{t^2/2}$ for all real $t$, that is, iff
\begin{equation*}
p\le p_u:=\inf_{t\in\R}r_u(t)=\inf_{t\ge0}r_u(t) \tag{3}
\end{equation*}
where $\ch:=\cosh$ and
\begin{equation*}
r_u(t):=\frac{e^{t^2/2}-1}{\ch tu-1}
\end{equation*}
for real $t\ne0$, with $r_u(0):=r_u(0+)=r_u(0-)=1/u^2$. Since $r_u(\infty-)=\infty$ and the function $r_u$ is even and continuous, the infimum in (3) is attained at some $t_u\in[0,\infty)$, so that
\begin{equation*}
p_u=r_u(t_u). \tag{4}
\end{equation*}

Since the set $[0,\infty]$ is compact, wlog one of the following cases must occur:

*Case 1: $t_u=0$.* Then
\begin{equation}
p_u=r_u(0)=\frac1{u^2}>>e^{-u^2/2}. \tag{5}
\end{equation}
Here and in the sequel, we write $a>>b$ for $a/b\to\infty$.

*Case 2: $t_u\downarrow0$ and even $t_u u\to0$.* Then
\begin{equation}
p_u=\frac{e^{t_u^2/2}-1}{\ch t_u u-1}\sim\frac1{u^2}>>e^{-u^2/2}. \tag{6}
\end{equation}

*Case 3: $t_u\downarrow0$ and $t_u u\to c$ for some $c\in(0,\infty)$.* Then $t_u\sim c/u$ and hence
\begin{equation}
p_u=\frac{e^{t_u^2/2}-1}{\ch t_u u-1}\sim\frac{t_u^2/2}{\ch c-1}
\sim\frac{c^2/2}{\ch c-1}\frac1{u^2}>>e^{-u^2/2}. \tag{7}
\end{equation}

*Case 4: $t_u\downarrow0$ and $v:=t_u u\to\infty$.* Then $v=o(u)$ and hence
\begin{equation}
p_u=\frac{e^{t_u^2/2}-1}{\ch t_u u-1}\sim\frac{v^2/2}{e^v/2}\frac1{u^2}
>>\frac{e^{-v}}{u^2}>>e^{-u^2/2}. \tag{8}
\end{equation}

*Case 5: $t_u\to c$ for some $c\in(0,\infty)$.* Then
\begin{equation}
p_u=\frac{e^{t_u^2/2}-1}{\ch t_u u-1}\sim\frac{e^{c^2/2}-1}{e^{cu}/2}>>e^{-u^2/2}. \tag{9}
\end{equation}

*Case 6: $t_u\to\infty$.* Then
\begin{equation}
p_u=\frac{e^{t_u^2/2}-1}{\ch t_u u-1}\sim\frac{e^{t_u^2/2}}{e^{t_u u}/2}\ge2e^{-u^2/2}. \tag{10}
\end{equation}

Thus, in all cases
\begin{equation}
p_u\ge(2-o(1))e^{-u^2/2}. \tag{11}
\end{equation}
Also, if we choose $p=p_u$, then clearly the non-strict inequality in (3) will hold. So, $X_{p_u,u}\in S$ and hence, by (2) and (2.5),
\begin{equation*}
P_u\ge P(X_{p_u,u}\ge u)=p_u/2.
\end{equation*}
Now (2) follows by (11). $\Box$