Let $a\_i (i \in\{1...k\})$ be $k$ IID standard Gaussian random variables, $P\_i$ are $d$-dimensional constant vectors. How to prove with a probability of at least $1/e$,
$$
\left\|\sum\_{i=1}^k a\_{i} P\_{i}\right\|\_2 \geq\left\|P\_{1}\right\|\_2 .
$$
holds.



prove with a probability of at least $1/e$: $\left\|\sum_{i=1}^k a_{i} P_{i}\right\|_2 \geq\left\|P_{1}\right\|_2$ holds
$\newcommand\si\sigma$Let
$$p:=P\Big(\Big\|\sum\_{i=1}^k a\_i P\_i\Big\| \ge\|P\_1\|\Big).$$
Here $\|\cdot\|:=\|\cdot\|\_2$.


The inequality $p\ge1/e=0.367\dots$ does not hold in general.


Actually, the best lower bound on the probability $p$  

is  

$$p\_\*:=P(|a\_1|\ge1)=0.317\dots,$$
which is $<1/e=0.367\dots$.


Indeed, on the one hand, if $P\_1\ne0=P\_2=\cdots=P\_k$, then clearly $p=p\_\*$. So, there is no lower bound on $p$ that is $>p\_\*$.


On the other hand, there is a unit $d$-dimensional vector $u$ such that $u\cdot P\_1=\|P\_1\|$, where $\cdot$ denotes the dot product. Then
$$\Big\|\sum\_{i=1}^k a\_i P\_i\Big\|\ge\Big|u\cdot\sum\_{i=1}^k a\_i P\_i\Big|
=\Big|\sum\_{i=1}^k c\_i a\_i\Big|=|X|, \tag{1}\label{1}$$
where $c\_i:=u\cdot P\_i$ and $X:=\sum\_{i=1}^k c\_i a\_i$, so that $c\_1=\|P\_1\|$. Without loss of generality (wlog), $\|P\_1\|\ne0$ -- otherwise, $p=1$. So, by rescaling, wlog $\|P\_1\|=1$ and hence $c\_1=1$. So, $X$ is a Gaussian random variable with mean $0$ and variance
$$\si^2:=\sum\_{i=1}^k c\_i^2=1+\sum\_{i=2}^k c\_i^2\ge1.$$
So, in view of \eqref{1},
$$p\ge P(|X|\ge1)=P(\sqrt{\si^2}\,|a\_1|\ge1)\ge P(|a\_1|\ge1)=p\_\*.$$


This concludes the proof that $p\_\*$ is the best lower bound on $p$. $\quad\Box$


