Determining the dual problem for the primal $\min_{x,t} t$ subject to $x_i\ge1$, $\|x\|_2\le t$, and $x_1+x_2=5$ I want to figure out the dual problem for the following primal
\begin{alignat*}{2}
&\underset{x,t}{\text{mininimize}}   &\qquad& t \\
&\text{subject to} &      & x_1 \geq 1 \\ 
& & & x_2 \geq 1 \\
& & & \vert|x\vert|_2 \leq t \\
& & & x_1+x_2 = 5
\end{alignat*}
I found the Lagrangian
$$L(x,t,\lambda_1,\lambda_2,\lambda_3,v) = t(1-\lambda_3)+x_1(v-\lambda_1)+x_2(v-\lambda_2)+\lambda_3 \vert|x\vert|_2 + \lambda_1+\lambda_2-5v.$$
The terms in $t$ are unbounded below unless $\lambda_3 = 1$. To get the dual we  need to solve
$$\min_{x} \left\{x_1(v-\lambda_1)+x_2(v-\lambda_2)+\vert|x\vert|_2\right\}.$$
How does one determine the solution? It seems like a lot of conditions would be required to guarantee this is unbounded in $x$.
 A: For simplicity, I will introduce the vector $\vec{z} = (v-\lambda_1,v-\lambda_2)$ so that we may write the optimization problem in question as
$$
f(\vec{z})=\min_{\vec{x}} \{\langle \vec{z},\vec{x}\rangle + \lVert\vec{x}\rVert_2 \}
$$
where $\langle\cdot,\cdot\rangle$ denotes the inner product of the vectors.
If it holds that $\lVert \vec{z}\rVert_2>1$, one may choose $\vec{x}=-\alpha\vec{z}$ for any value $\alpha>0$ such that
\begin{align*}
\langle \vec{z},\vec{x}\rangle + \lVert\vec{x}\rVert_2 
&= -\alpha\lVert\vec{z}\rVert_2^2 + \alpha \lVert\vec{z}\rVert_2 \\
&= \alpha\lVert\vec{z}\rVert_2 \left(1 - \lVert\vec{z}\rVert_2\right),
\end{align*}
and thus the minimization problem is unbounded from below. On the other hand, if it holds that $\lVert \vec{z}\rVert_2\leq 1$ then 
$$
\langle \vec{z},\vec{x}\rangle \geq - \lVert \vec{x}\lVert_2
$$
holds for any $\vec{x}$. 
We may therefore conclude that 
$$
f(\vec{z}) = \left\{ \begin{array}{ll} 0 & \text{if }\lVert \vec{z}\rVert_2\leq 1\\-\infty & \text{if } \lVert \vec{z}\rVert_2>1.\end{array}\right.
$$
The dual problem can thus be stated as follows:
\begin{alignat*}{2}
&\underset{v,\lambda_1,\lambda_2}{\text{maximize}}   &\qquad& \lambda_1 + \lambda_2 -5v \\
&\text{subject to} &      & \left\lVert \begin{pmatrix}v-\lambda_1\\v-\lambda_2\end{pmatrix}\right\rVert_2\leq1\\
& & & \lambda_1,\lambda_2\geq0
\end{alignat*}
