Bound on the shortest non-zero vector in any full rank n-dimensional lattice $\Lambda \subseteq \mathbb{R}^n$ with respect to the $1$-norm. How can i prove
$$\lambda_1 \; \leq \; (n! \; det(\Lambda))^{\frac{1}{n}} \approx \frac{n \; det{(\Lambda)}^\frac{1}{n}}{e}.$$
Here $\Lambda_1$ is shortest non-zero vector. My initial thought was using Minkowski theorem (choosing $S =$ n-Ball of radius $\sqrt n \frac{\lambda_1}{n}$) and proof by contradiction (assuming $\lambda_1 \; > \; (n! \; det(\Lambda))^{\frac{1}{n}}$ and contracting with minimality of $\lambda_1$).
[Minkowski’s convex body theorem] : Let $\Lambda$ be a full dimensional lattice.  If $S \subset \mathbb{R}^n$ is a symmetric convex body of volume $vol(S) > 2^n det(\Lambda)$ ,  then $S$ contains a  non-zero lattice point.
$1-$norm for a vector $x$ : $\sum{}{}{|x_i|}$.
 A: $$C_n=Vol(\{ x, \|x\|_1\le 1\})= \int_{-1}^1 Vol(\{ y, \|y\|_1\le 1-x_1\})dx_1$$ $$=\int_{-1}^1 (1-|x_1|)^{n-1}C_{n-1}dx_1=\frac{2}n C_{n-1}=\frac{2^n}{n!}$$
For all $r<\lambda_1/2$ then $ \{x, \|x\|_1\le r\}$ doesn't intersect its $\Lambda$-translates. Thus $$\det(\Lambda)\ge Vol(\{ x, \|x\|_1\le r\})=r^n\frac{2^n}{n!}$$ and hence $$\lambda_1^n\le  n! \det(\Lambda)$$
A: I would like to add my answer too, somebody may find it useful. We show that every (full rank, n-dimensional) lattice $\Lambda$ always contains a non-zero vector $x$ such that
$||x||_1 \leq \; (n! \; det(\Lambda))^{\frac{1}{n}}$.
Let
$\lambda_1 = min\{||x||_1 : x \in \Lambda \setminus \{0\}\}$ and assume for contradiction $l > (n! \; det(\Lambda))^{\frac{1}{n}}$. As @reuns  mentioned take
$S = \{ x : ||x||_1 < l \}$. Notice that $S$ is convex, symmetric and has volume
$vol(S) = 2^n \frac{l^n}{n!} > 2^n det(\Lambda)$. So, by Minkowski's theorem, $S$ contains a non-zero lattice vector $x$. By definition of $S$ ,we have$||x||_1 < l$, a contradiction to the minimality of $l$.
