LLL guarantees that we can find a basis $v\_1,\dots,v\_n$ of a lattice in $\mathbb R^n$ with


$$\|v\_i\|\leq \gamma\_{i,n} \det(\Lambda)^{1/(n-i+1)}$$ where $\gamma\_i$ is a function only of $i$ and $n$.


1. Are there lattices where this cannot be improved to $\|v\_i\|\leq \gamma\_{i,n} \det(\Lambda)^{1/(n-i+2)}$?
2. In general are there algorithms (possibly in exponential time) which can guarantee $\|v\_i\|\leq \gamma\_{i,n} \det(\Lambda)^{1/(n-i+2)}$?



Lowering $i$th shortest vector of a lattice
There are lattices where your requirement cannot be met. In fact if you fix any lattice $\Lambda\_0$ and you dilate it by a factor of $R$, then its determinant gets scaled by $R^n$, while its shortest lattice vector gets scaled by $R$. However, your inequality $\|v\_1\|\ll\_n\det(\Lambda)^{1/(n+1)}$ would yield that the shortest lattice length grows at most at the rate of $R^{n/(n+1)}$, which is a contradiction.


You can get further lower bounds on the lengths of the basis vectors by using Minkowski's theorem that the successive minima of $\Lambda$ satisfy 
$$\lambda\_1\lambda\_2\dots\lambda\_n\asymp\_n\det(\Lambda).$$ 
Note also that the reduced basis produced by the LLL algorithm is optimal in the sense that the length of the $i$-th basis vector is $\asymp\_n\lambda\_i$. This is explicitly mentioned in the original paper of Lenstra-Lenstra-Lovász (1982), see their remark below (1.13).


