Trace inequality If $$\sum \Vert Ae_n \Vert \leq \sum \Vert B e_n \Vert$$
How we can conclude that
$$\operatorname{Tr}|A| \leq \operatorname{Tr}|B|$$
where $|A|=(A^{\ast} A)^{1/2}$ and $e_n$ are orthonormal basis
 A: In reading your question I presume $A^*$ indicates a conjugate transpose and by $||\vec v||$ you mean the norm $\sqrt{\vec v\cdot\vec v}$ where $\vec v\cdot\vec v\equiv\vec v^*\vec v$. I'll use the same convention.
Let $\vec a_n$ be the columns of $A$. Then the left side of your first inequality can be written as:
$$\sum ||A\hat e_n||=\sum||\vec a_n||=\sum\sqrt{\vec a_n\cdot\vec a_n}$$
It follows from the properties of matrix multiplication that the entries of $A^*A$ are dot products of its columns. Specifically:
$$(A^*A)_{ij}=\vec a_i\cdot\vec a_j$$
If we take the trace, we sum over indices
$$\text{Tr}(A^*A)=(A^*A)_{1,1}+(A^*A)_{2,2}+...=\sum(A^*A)_{nn}=\sum\vec a_n\cdot\vec a_n$$
Now if we let $\sqrt{\ }$ denote an elementwise square root, we can take the trace of $\sqrt{A^*A}$ by taking the square root of the elements in the sum.
$$\text{Tr}\sqrt{A^*A}=\sum\sqrt{\vec a_n\cdot\vec a_n}$$
From this, we see that the quantities on the left sides of both inequalities are identical. Clearly, the same is true for the right sides. Therefore your inequalities describe exactly the same statement, constructed in slightly different ways.
If, instead, we interpret $\sqrt{}$ as a matrix square root (that is, $\sqrt A$ is a unique matrix s.t. $\sqrt A\sqrt A=A$, then the the inequalities are not equivalent. To see why notice that the statement can be rephrased as follows.
Let $A$ and $B$ be positive semidefinite Hermitian matrices and $\sqrt A$, $\sqrt B$ be their matrix square roots.
$$\sum\sqrt{A_{ii}}\le\sum\sqrt{B_{ii}}\implies\text{Tr}\sqrt A\le\text{Tr}\sqrt{B}$$
A counterexample to this would be
$$A=\begin{bmatrix}
1        & \frac 12 \\
\frac 12 & 1
\end{bmatrix},\ \ \ B=\begin{bmatrix}
1        & \frac 16 \\
\frac 16 & 1
\end{bmatrix}$$
