How to prove that the second optimization problem is the dual problem of the first problem. I tried to prove that using Lagrangian dual function.
First optimization problem:
$$\begin{array}{ll} \text{minimize} & x^TQx\\ \text{subject to} & \\ \qquad e^Tx = 1\\ \qquad \mu^Tx \ge R\\ \qquad x \ge 0\\ \qquad e = [1,1,...,1]^T \end{array}$$
Second optimiztaion problem:
$$\begin{array}{ll} \text{maximize} & \mu^Tx\\ \text{subject to} & \\ \qquad e^Tx = 1\\ \qquad x^TQx \le \sigma\\ \qquad x \ge 0\\ \qquad e = [1,1,...,1]^T \end{array}$$
where Q is an n by n positive definite matrix. $x \in R^n,\ \mu \in R^n$ . $\ x$ and $\mu$ are n dimensional column vectors.
I tried to solve this by using Lagrangian dual function:
$$ L(x,\lambda,\alpha,\beta) = x^TQx+\lambda(R-\mu^Tx)-\alpha^Tx+\beta(e^Tx-1)=x^TQx+(\beta e - \alpha - \lambda \mu)^T x + \lambda R - \beta\\ \frac{\partial{L(x,\lambda,\alpha,\beta)}}{\partial{x}}=2Qx+\beta e-\alpha-\lambda \mu = 0\\ x = \frac{1}{2}Q^{-1}(\alpha+\lambda \mu-\beta e) $$
Suppose t = -x, substitute t for x and take maximize, then we have the dual problem:
$$ \max_{\lambda \ge 0,\alpha \ge 0,\beta } -t^TQt+\lambda R - \beta\\ s.t. \quad t = \frac{1}{2}Q^{-1}(\beta e - \alpha - \lambda \mu) $$ Then I don't know how to continue to tranform this problem into the form of the second optimization problem.