How tight is the bound $P(\|X\|^2 \ge t |\langle a,X\rangle|) \ge 1 - t\sqrt{\frac{2}{m-1}}$, where $X \sim N(0, I_m)$ and $\|a\| = 1$? Let $X$ be a random vector in $\mathbb R^m$ with iid $N(0,1)$ coordinates and let $a$ be a fixed unit vector in $\mathbb R^m$. In another post (SE link here https://math.stackexchange.com/a/3792730/168758),  I'm able to show using anti-concentration results for quadratic forms in gaussians, that
$$
P(\|X\|^2 \ge t |\langle a,X\rangle|) \ge 1 - t\sqrt{\frac{2}{m-1}}.
$$

Question. Can the above bound be improved ?

 A: $\newcommand\Ga\Gamma$
$\newcommand{\R}{\mathbb R}$
This bound can be greatly improved, even using rather rough estimates.
Indeed, in view of the spherical symmetry of the distribution of $X$, without loss of generality $a$ is the first vector of the standard basis of $\R^m$. So, your inequality can be rewritten as
\begin{equation*}
    p_n(t):=P(W<t|Z_1|)\le t\sqrt{\frac2n}=:r_n(t), \label{1}\tag{1}
\end{equation*}
where $t>0$, $W:=Z_1^2+\dots+Z_{n+1}^2$, the $Z_i$'s are iid standard normal, and
\begin{equation*}
    n:=m-1\ge1. 
\end{equation*}
Note that for $t\ge\sqrt{n/2}$ inequality \eqref{1} is trivial. So, without loss of generality
\begin{equation*}
    0<t<\sqrt{n/2}. \label{2}\tag{2}
\end{equation*}
Let us give an upper bound on $p_n(t)$ which is much better than $r_n(t)$. Letting $Y:=Z_2^2+\dots+Z_{n+1}^2$, we have
\begin{equation*}
    p_n(t)=P(Y<t|Z_1|-Z_1^2)=\int_0^\infty P(Y<y)P(t|Z_1|-Z_1^2\in dy). 
\end{equation*}
Since $Y$ has the gamma distribution with parameters $n/2$ and $2$, for $y>0$ we have
\begin{equation*}
    P(Y<y)=c_n\int_0^y u^{n/2-1}e^{-u}\,du\le c_n y^{n/2},
\end{equation*}
where
\begin{equation*}
    c_n:=\frac1{\Ga(n/2)2^{n/2}}. 
\end{equation*}
Thus,
\begin{equation*}
    p_n(t)\le c_ne_n(t),
\end{equation*}
where
\begin{align*}
    e_n(t)&:=\int_0^\infty y^{n/2}P(t|Z_1|-Z_1^2\in dy) \\ 
    &=E\max(0,t|Z_1|-Z_1^2)^{n/2} \\ 
    &=\frac1{\sqrt{2\pi}}\int_{-t}^t (t|z|-z^2)^{n/2}e^{-z^2/2}\,dz \\ 
    &=\frac2{\sqrt{2\pi}}\int_0^t (tz-z^2)^{n/2}e^{-z^2/2}\,dz \\ 
    &\le\frac2{\sqrt{2\pi}}\int_0^t (tz-z^2)^{n/2}\,dz \\ 
&=\frac{2t^{n+1}}{\sqrt{2\pi}}\int_0^1 (u-u^2)^{n/2}\,du \\ 
&=\frac{2t^{n+1}}{\sqrt{2\pi}}\frac{\Ga(n/2+1)^2}{\Ga(n+2)}=:f_n(t).  
\end{align*}
We conclude that
\begin{equation*}
    p_n(t)\le q_n(t):=c_n f_n(t). 
\end{equation*}
The ratio of the new bound, $q_n(t)$, to the old bound, $r_n(t)$, is
\begin{equation*}
    R_n(t):=\frac{q_n(t)}{r_n(t)}=\frac{t^n\sqrt n}{\sqrt{\pi}}\frac{\Ga(n/2+1)^2}{\Ga(n+2)}\frac1{\Ga(n/2)2^{n/2}}
=\frac{t^n n^{3/2}}{\sqrt{\pi}}\frac{\Ga(n/2+1)}{\Ga(n+2)}\frac1{2^{n/2+1}}
    \le R_n(\sqrt{n/2}),
\end{equation*}
in view of \eqref{2}. Also, by Stirling's formula,
\begin{equation*}
    R_n(t)\le R_n(\sqrt{n/2})\sim\frac{\sqrt{2n/\pi}}4\,\Big(\frac e8\Big)^{n/2}
\end{equation*}
as $n\to\infty$.
So, the new bound is exponentially smaller than the old bound uniformly over $t$ as in \eqref{2}, and the improvement is only much greater for $t=o(\sqrt n)$.

Here are the graphs $\{(t,R_n(t))\colon0<t<\sqrt{n/2}\}$ of the ratios of the new upper bound, $q_n(t)$, to the old bound, $r_n(t)$, for $n=10$ (left) and $n=20$ (right).

