Find the minimizer of $w_1 \|x - a\|_1 + w_2 \|x - b\|_2$ 
Find $$\arg\min_{x \in \mathbb{R}^n} w_1\|x - a\|_1 + w_2\|x - b\|_2$$

I'm trying to evaluate 
$$\hat{x} := \arg\min_{x \in \mathbb{R}^n} w_1\|x - a\|_1 + w_2\|x - b\|_2 \tag{1}$$ 
to find a closed form, or at least a simpler expression, in terms of $w_1 > 0, w_2 > 0, a \in \mathbb{R}^n$, and $b \in \mathbb{R}^n$. While this is vague, I'm hoping to have an expression that is reasonably interpretable.
Solution attempt I
In my first attempt at finding a closed form, I took the subgradient, which may find a candidate minimizer. From the first order condition, we know that $\hat{x}$ satisfies that the subgradient of the objective in (1) is zero; that is, $$0 = w_1 \hat{z} + w_2  \frac{\hat{x}-b}{\|\hat{x}-b\|_2},$$ where 
$$\hat{z} \in \begin{cases} \{\mathrm{sgn}(\hat{x} - a)\} & \mathrm{ if } \hat{x} \ne a \\ [-1,1] & \mathrm{ if } \hat{x} = a \end{cases}$$ 
While this is a characterization of $\hat{x}$, it isn't clear to me how to proceed to find a closed form.
Solution attempt II
As a second attempt, I tried to use duality to find a simpler expression.
By introducing $y = x-a$, we can see that this problem is equivalent to $$\hat{y} = \arg\min_{y \in \mathbb{R}^n} w \|y\|_1 + \|y - (b-a)\|_2,$$ where $w = \frac{w_1}{w_2}$. By the Lagrangian duality, we know that there exists some $C \in \mathbb{R}$ so that 
\begin{align}
  \hat{y} 
  & = \arg\min_{\|y\|_1 \le C} \|y - (b-a)\|_2 \\
  & = \arg\min_{\|y\|_1 \le C} \|y - (b-a)\|_2^2 \\
  & = \arg\min_{y} \|y - (b-a)\|_2^2 + \lambda \|y\|_1 \\
  & = \mathrm{sgn}(b-a) \left( |b-a| - \lambda \right)_+,
\end{align}
is just soft thresholding, for some dual variable $\lambda$ that depends on $C$ and $b-a$ in some way that I don't understand. This appears to be a closed form, but, since $\lambda$ is not a closed form function of $w_1, w_2, a$ and $b$, this doesn't satisfy what I'm looking for.
 A: I'll solve it first for the case of $\mathbb{R}^1$.  Then I’ll use that intuition to solve it generally.
Write your objective function as $f(x)$.  It is real valued, and it is smooth everywhere but $a=x$ and $x=b$.  I'll use the notation that $\hat{x}$ is a minimizer.
If $a=b$, then the trivial solution is $\hat{x}=a=b$. So assume $a \ne b$.  I am also going to assume that $w_1 >0$ and $w_2 > 0$; otherwise the solution is again trivial. 
In this easy case, your objective function is simply
$$
f(x)=w_1|x-a|+ w_2\sqrt{(b-x)^2}
$$
This function is minimized by setting $\hat{x}=a$ if $w_1 \ge w_2$ and $\hat{x}=b$ otherwise.  (If $w_2=w_1$, then any $x$  between $a$ and $b$ is a minimizer.)
I will now try to extend this solution to $\mathbb{R}^n$.  
The objective function $f(x)$ is continuous, and any bounded $X \subset \mathbb{R}^n$ that contains $a$ and $b$ is compact.  So $f(x)$ attains a minimum on $X$. We know that the minimizer $\hat{x}$ will satisfy
$$
f(\hat{x}) \le min\{f(a),f(b) \}= min \{w_1 \parallel a- b \parallel_1,w_2\parallel a- b \parallel_2 \}
$$
Write $I_i=[min\{a_i,b_i\} , max\{a_i,b_i\}]$.  Notice that $\hat{x}_i \in I_i$  because otherwise the objective function could be made lower by moving $\hat{x}_i$ into that closed interval. Now put $X=I_1 \times \cdots \times I_n$.  From now on, we will restrict $f(.)$ to this compact domain.  Let $U \subset \mathbb{R}^n$ be an open set containing $X$.  Note further that $f(x)$ is smooth everywhere in $U$, except at $x_i = a_i$ or $x = b$.  So $f(.)$ is smooth, even on the boundary of $X$, except at those values.
The first order necessary condition with respect to $x_i$ is 
$$
\begin{array}{c}
\partial f / \partial x_i = \pm  w_1 - w_2 \dfrac{b_i - \hat{x}_i}{\parallel b-\hat{x} \parallel_2}= 0 \\
\pm  w_1 = w_2 \dfrac{b_i - \hat{x}_i}{\parallel b- \hat{x} \parallel_2}.
\end{array}
$$
since $w_1 > 0$ and $w_2 > 0$, none of the $\hat{x}_i=b_i$ unless $\hat{x}=b$. Summing the squares, we derive:
$$
n  w_1^2 = w_2^2  \sum_i \dfrac{(b_i - \hat{x}_i)^2}{\parallel b-\hat{x} \parallel_2^2} \\
n w_1^2=w_2^2
$$ 
which cannot be true in general since $w_1$ and $w_2$ are arbitrary.  If it were true, then any $\hat{x}$ in the interior of $X$ would be a minimizer. 
We have shown that $\hat{x_i}=a_i$ for some $i$ or $\hat{x}=b$, except for the special case when $nw_1^2=w_2^2$.
If $\hat{x}_i=a_i$, then moving away from $a_i$ must not decrease the objective function.  Thus:
$$
w_1^2 \ge w_2^2 \dfrac{(b_i-a_i)^2}{\parallel b-\hat{x} \parallel^2_2},
$$ 
with equality for the indices for which $\hat{x}_j \ne a_j$.  Squaring and summing again, we derive:
$$
nw_1^2 \ge w_2^2
$$
Hence, if $w_1$ is sufficiently large, then there is some $x_i=a_i$.  This is the appropriate generalization of the condition $w_1 \ge w_2$ for the case where $n=1$.
We can say more.  Let 
$$
\begin{array}{c}
m_i=\dfrac{(b_i-a_i)^2}{\parallel b- a \parallel_2^2} \\
\end{array}
$$
and notice that 
$$
\begin{array}{c}
m_i < \dfrac{(b_i-a_i)^2}{\parallel b - \hat{x} \parallel_2^2} \\
\end{array}
$$ 
for any $\hat{x} \ne a$.  Then write $m = min \{m_1, ..., m_n\}$ and $M=max \{m_1, ..., m_n\}$.   If $m_i \le m_k$ and $\hat{x}_k =a_k$, then we know that $\hat{x}_i=a_i$ too.  So if $w_1 \ge M w_2$ then $\hat{x}=a$.  And if $w_1 < m w_2$, then $\hat{x}_i \ne a_i$ for any $i$.  
If  $w_1 < m w_2$, then the least squares formula gives $\hat{x}$.  Let $y=sign(a_i-b_i)w_1/w_2$.  Then
$$
\hat{x} = b(b^Tb)^{-1}b^Ty 
$$
Finally, if $\hat{x}=b$, then moving away from any $b_i$ must not decrease the objective function. We need only consider $x_i=a_i$.  Now we cannot use the first-order condition.  Hence
$$
w_1^2 \le w_2^2 (b_i-a_i)^2
$$
for every $i$. Summing again, we see:
$$
nw_1^2 \le w_2^2 \parallel b-a \parallel_2^2
$$
So for low penalty $w_1$, we put $\hat{x} = b$
A: Given vectors $\mathrm a, \mathrm b \in \mathbb R^n$ and weights $w_1, w_2 > 0$
$$\text{minimize} \quad w_1 \| \mathrm x - \mathrm a \|_1 + w_2 \| \mathrm x - \mathrm b \|_2$$
Introducing optimization variables $\mathrm y \in \mathbb R^n$ and $z \in \mathbb R$, I believe the unconstrained optimization problem above can be rewritten as the following constrained optimization problem
$$\begin{array}{ll} \text{minimize} & w_1 \, 1_n^\top \mathrm y + w_2 \, z\\ \text{subject to} & -\mathrm y \leq \mathrm x - \mathrm a \leq \mathrm y\\ & \| \mathrm x - \mathrm b \|_2 \leq z\end{array}$$
Squaring both sides of $\| \mathrm x - \mathrm b \|_2 \leq z$, we obtain
$$\left( \mathrm x - \mathrm b \right)^\top \left( \mathrm x - \mathrm b \right) \leq z^2$$
Dividing both sides by $z > 0$, we obtain
$$\left( \mathrm x - \mathrm b \right)^\top \left( z \mathrm I_n \right)^{-1} \left( \mathrm x - \mathrm b \right) \leq z$$
Using the Schur complement, we obtain the following linear matrix inequality (LMI)
$$\begin{bmatrix} z \mathrm I_n & \left( \mathrm x - \mathrm b \right)\\ \left( \mathrm x - \mathrm b \right)^\top & z\end{bmatrix} \succeq \mathrm O_{n+1}$$
Thus, we have a semidefinite program (SDP) in $\mathrm x, \mathrm y \in \mathbb R^n$ and $z \in \mathbb R$
$$\boxed{\begin{array}{ll} \text{minimize} & w_1 \, 1_n^\top \mathrm y + w_2 \, z\\ \text{subject to} & -\mathrm y \leq \mathrm x - \mathrm a \leq \mathrm y\\ & \begin{bmatrix} z \mathrm I_n & \left( \mathrm x - \mathrm b \right)\\ \left( \mathrm x - \mathrm b \right)^\top & z\end{bmatrix} \succeq \mathrm O_{n+1}\end{array}}$$
