We want to minimize $x_1+x_2$ and $x_1,x_2 \ge 0$, so we have to get close to origin as much as possible. Consider the problems in $x_1-x_2$ plane, first we have to find the feasible region (I am assuming $a_1,a_2\ge0$):
The blue curve is the boundary of the first inequality $x_1 x_2 \ge a_1$, the feasible region is the region above this curve. The dashed-line orange curves are the second inequality $x_1x_2\ge ya_2$ for different values of $y$, again the area above these curves are the feasible region. This means if $y\le\frac{a_1}{a_2}$, we can ignore the second inequality, otherwise (if $y\ge\frac{a_1}{a_2}$) we can ignore the first inequality. The purple vector shows the direction of movement of curves as we increase $y$. Then we have the third inequality $x_1 \ge y$ which has the boundary of green line, our answer is in the right hand side of green line. And finally we have the forth inequality $x_2 \le y$ with boundary of grey line, our answer is in lower half of this line (below grey line).
With these information at hand, we see that the inequality $x_1\ge y$ must become active at solution point, this means the slackness condition $\lambda_4(y-x_1)=0$ is equivalent to $x_1=y$. Substituting this in the primal, we get three inequalities $x_2 \ge \frac{a_1}{y}$, $y\ge x_2$ and $x_2 \ge a_2$ and the objective is $y+x_2=x_1+x_2$.
Back to primal. Now consider on one hand we have we have $y\ge x_2$ and $x_2 \ge \frac{a_1}{y}$ which means $y \ge \frac{a_1}{y}$ or $y \ge \sqrt{a_1}$. On the other hand we have $y\ge x_2$ and $x_2 \ge a_2$which means $y\ge a_2$. Thus we obtain two main conditions that solves everything: $y \ge \sqrt{a_1}$ and $y\ge a_2$.
Finally if $a_2 \le \sqrt{a_1}$ the solution is $x_1=y=x_2=\sqrt{a_1}$. Otherwise if $\sqrt{a_1} \le a_2$ the solution is $x_1=x_2=y=a_2$. And now we can see the funny part, both $x_1 \ge y$ and $y \ge x_2$ are tight so from start we could have considered the slackness conditions $\lambda_3(y-x_1)=0$ and $\lambda_4(x_2-y)=0$ to be active i.e. $x_1=x_2=y$ to be true and get the solution.
We define the Lagrangian
$$
\mathcal{L}(x_1, x_2, \lambda_1, \lambda_2) = f_0(x_1, x_2) - \lambda_1 f_1(x_1, x_2) - \lambda_2 f_2(x_1, x_2) - \lambda_3 f_3(x_1, x_2)
$$
Suppose now that all constraints are active. Under Slater's, the 1st order necessary KKT conditions state that
$$
\begin{align}
\partial_{x_1} \mathcal{L} = 0 \\
\partial_{x_2} \mathcal{L} = 0 \\
\partial_{\lambda_1} \mathcal{L} = 0 \\
\partial_{\lambda_2} \mathcal{L} = 0 \\
\partial_{\lambda_3} \mathcal{L} = 0
\end{align}
$$
Now let's see what the solutions to this would look like. These are stated as
\begin{align}
4 x_1 - 15 - 4 x_2 - \lambda_1 + \lambda_3 = 0 \\
8 x_2 -30 - 4 x_1 - \lambda_2 + 2\lambda_3 = 0 \\
x_1 = 0 \\
x_2 = 0 \\
30 - x_1 - 2x_2 = 0
\end{align}
but clearly the three last equations cannot be satisfied. Thus the problem is not minimized with three active constraint.
Suppose now only two constraints are satisfied. Considering first constraint 1 and 2, this yields
\begin{align}
4 x_1 - 15 - 4 x_2 - \lambda_1 = 0 \\
8 x_2 -30 - 4 x_1 - \lambda_2 = 0 \\
x_1 = 0 \\
x_2 = 0 \\
\end{align}
which is a linear system and is satisfied iff $x_1 = 0, x_2 = 0, \lambda_1 = -15, \lambda_2 = -30$. But Lagrange multipliers should be positive.
We continue with constraint 1 and 3
\begin{align}
4 x_1 - 15 - 4 x_2 - \lambda_1 + \lambda_3 = 0 \\
8 x_2 -30 - 4 x_1 + 2\lambda_3 = 0 \\
x_1 = 0 \\
30 - x_1 - 2x_2 = 0
\end{align}
which is satisfied for $x_1 = 0, x_2 = 15, \lambda_3 = 75, \dots$.
$$
f(0, 15) = 450 > f(1, 1) = -43
$$
So this is not our minimizer
We continue with constraint 2 and 3
\begin{align}
4 x_1 - 15 - 4 x_2 + \lambda_3 = 0 \\
8 x_2 -30 - 4 x_1 - \lambda_2 + 2\lambda_3 = 0 \\
x_2 = 0 \\
30 - x_1 - 2x_2 = 0
\end{align}
which is satisfied for $x_1 = 30, x_2 = 0, \lambda_3 = -105$. As lagrange multiplier should be positive this is not our solution either.
Suppose now only one constraint is active,
\begin{align}
4 x_1 - 15 - 4 x_2 - \lambda_1 = 0 \\
8 x_2 -30 - 4 x_1 = 0 \\
x_1 = 0 \\
\end{align}
which yields $x_1 = 0, x_2 = 15/4, \lambda_1 = -75/4$. As before this is not our solution
OR
\begin{align}
4 x_1 - 15 - 4 x_2 = 0 \\
8 x_2 -30 - 4 x_1 - \lambda_2 = 0 \\
x_2 = 0 \\
\end{align}
which yields $x_1 = 15/4, x_2 = 0, \lambda_2 = -45$. As before this is not our solution
OR
\begin{align}
4 x_1 - 15 - 4 x_2 + \lambda_3 = 0 \\
8 x_2 -30 - 4 x_1 + 2\lambda_3 = 0 \\
30 - x_1 - 2x_2 = 0
\end{align}
which yields $x_1 = 12, x_2 = 9, \lambda_3 = 3$. We inspect
$$
f(12, 9) = -270
$$
So clearly a good candidate.
Finally we check that the problem is not solved for all constraints inactive,
\begin{align}
4 x_1 - 15 - 4 x_2 = 0 \\
8 x_2 -30 - 4 x_1 = 0 \\
\end{align}
which yields $x_1 = 15, x_2 = 45/4$. However this solution is not satisfied by the constraints.
To summarize: We considered possible combinations of active constraints and solved the KKT conditions for the (1 + 3 + 3) = 7 different scenario. In all the cases the KKT conditions are linear systems. In the 7 different scenario there are only 2 candidates, $(0, 15)$ and $(12, 9)$. The value attained at $(12, 9)$ is the smallest and we thus conclude that this is the solution to the problem. Wolfram Alpha agrees with me.
Best Answer
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*}