How Hermite Lattice reduced basis implies LLL reduced basis?
Ignoring the trivial case $j = i$, for which the second condition is automatically satisfied, the conditions for Hermite-reduced bases can be formally written as:


For all $1 &lt; i \leq n$ and for all $1 \leq j &lt; i$: $ \qquad \quad \quad \ \ |\mu_{i,j}| \le \frac{1}{2}$.
For all $1 \leq i &lt; n$ and for all $i + 1 \ {\color{red}\leq} \ j \leq n$: $\ \quad \|\pi_i(b_i)\| \le \|\pi_i(b_j)\|$.


The conditions for LLL-reduced bases can be slightly rewritten as: 


For all $1 &lt; i \leq n$ and for all $1 \leq j &lt; i$: $\qquad \quad \quad \ \ |\mu_{i,j}| \le \frac{1}{2}$.
For all $1 \leq i &lt; n$ and for all $i + 1 \ {\color{red}=} \ j\leq n$: $\ \quad \|\pi_i(b_i)\| \le \|\pi_i(b_j)\|$ .


The only difference in the definitions, highlighted in red, is on the values $j$ for which the second condition must hold; for Hermite-reduced bases it must hold for all $j \in \{i+1, \dots, n\}$ while for LLL-reduced bases it must hold for $j \in \{i + 1\}$ (provided $i + 1 \leq n$). Since $\{i + 1\} \subset \{i+1, \dots, n\}$ this means Hermite-reduction is a stronger notion than LLL-reduction. So Hermite-reduction implies LLL-reduction.

As for your last questions:


  What about the other inequalities? Does other inequalities are satisfied? If not, how does it output nearly orthogonal basis?


For the case $n = 2$, the sets $\{i + 1, \dots, n\}$ and $\{i + 1\}$ are equal for $i = 1$, while for $i = 2 = n$ the second condition does not apply. So for $n = 2$, both notions of reducedness are equivalent. (Many other notions of reducedness also collapse to the same definition in case $n = 2$.)

For $n &gt; 2$, setting $i = 1$ shows that the conditions are not necessarily equivalent. I am not sure whether already for $n = 3$ there are examples of LLL-reduced bases which are not Hermite-reduced, but certainly for larger $n$ there are many such examples - by the second condition, the vector $b_1$ of a Hermite-reduced basis must be a shortest non-zero vector of the lattice, while for LLL-reduced bases the vector $b_1$ is only known to be at most a factor $2^{O(n)}$ longer than a shortest non-zero vector in the lattice. In high dimensions, an LLL-reduced basis will look quite skewed compared to a Hermite-reduced basis, so in that sense it does not really output a "nearly orthogonal basis" - only a "somewhat orthogonal basis".