Dual Problem Solution
The Lagrangian is given by:
$$ L \left( x, \lambda, \nu \right) = {x}^{T} x + {\lambda}^{T} \left( x - a \right) + \nu \left( \boldsymbol{1}^{T} x - b \right) = {x}^{T} x + \left(
\lambda + \nu \boldsymbol{1} \right)^{T} x -{\lambda}^{T} a - \nu b $$
The Dual Function is given by:
$$ g \left( \lambda, \nu \right) = \inf_{x} L \left( x, \lambda, \nu \right) $$
Looking at the term related to $ x $:
$$ \inf_{x} {x}^{T} x + \left( \lambda + \nu \boldsymbol{1} \right)^{T} x $$
Which is a quadratic form of $ x $ with its minimizer given by:
$$ {x}^{\ast} = -\frac{1}{2} \left(
\lambda + \nu \boldsymbol{1} \right) $$
Its minimum given by
$$ {{x}^{\ast}}^{T} {x}^{\ast} + \left(
\lambda + \nu \boldsymbol{1} \right)^{T} {x}^{\ast} = -\frac{1}{4} \left(
\lambda + \nu \boldsymbol{1} \right)^{T} \left( \lambda + \nu \boldsymbol{1} \right) $$
Hence the Dual Problem is given by:
\begin{align*}
\text{maximize} & \quad & -\frac{1}{4} \left(
\lambda + \nu \boldsymbol{1} \right)^{T} \left( \lambda + \nu \boldsymbol{1} \right) - {\lambda}^{T} a - \nu b \\
\text{subject to} & \quad & \lambda \succeq 0
\end{align*}
The problem is Concave in $ \left( \lambda, \nu \right) $ hence it is a convex problem.
It can be solved by a Quadratic Programming as:
$$ \left(
\lambda + \nu \boldsymbol{1} \right)^{T} \left( \lambda + \nu \boldsymbol{1} \right) = {\left\| E v \right\|}_{2}^{2} = {v}^{T} {E}^{T} E v = {v}^{T} H v $$
Where $ v = {\left[ \lambda, \nu \right]}^{T}, \; E = \left[ I, \boldsymbol{1} \right], \; H = {E}^{T} E $. Then the problem becomes:
$$
\begin{align*}
\text{minimize} & \quad & \frac{1}{4} {v}^{T} H v + {v}^{T} f \\
\text{subject to} & \quad & A v \preceq 0
\end{align*}
$$
Where $ A = - \left[ I, \boldsymbol{0} \right], \; f = {\left[ a, b \right]}^{T} $.
The above can directly solved by MATLAB's quadprog(). Then $ {x}^{\ast} = -0.5 E v $.
Best Answer
The problem is convex, as demonstrated by the following second-order cone formulation, obtained by introducing $y_i$ to represent $1/x_i$: \begin{align} &\text{minimize} &\sum_i c_i y_i \\ &\text{subject to} & A x &\le b \\ && z &= \sqrt 2 \\ && 2 x_i y_i &\ge z^2 &&\text{for all $i$} \tag1\label1\\ && x_i &\ge 0 &&\text{for all $i$}\\ && y_i &\ge 0 &&\text{for all $i$} \end{align} Constraint \eqref{1} is a rotated second-order cone constraint. Everything else is linear.