I am trying to solve the following Lagrangian but I am having a hard time to find the solution. My end goal is to obtain a formula for x with a closed form solution for the lambda's
$$ L(x, \lambda_1, \lambda_2) = E(R) x^T - \lambda_1(x \vec{1} -1) - \lambda_2(x\Sigma x' - \sigma^2)=0 $$
Wrt to $x$
$$ L_x = E(R) - \lambda_1 \vec{1} - 2 \lambda_2 x^T \Sigma = 0 \iff x = \frac{1}{2\lambda_2} (E(R) - \lambda_1 \vec{1}) \Sigma^{-1} $$
Next I can derive wrt to $\lambda_1$
$$ \begin{aligned} L_{\lambda_1} = x \vec{1}' -1 &= 0\\ x \vec{1}' &= 1 \\ \frac{1}{2\lambda_2} (E(R) - \lambda_1 \vec{1}) \Sigma^{-1} \vec{1}' &= 1 \\ \frac{1}{2} b -\lambda_1 a &= \lambda_2 \end{aligned} $$
My problem lies wrt to $\lambda_2$, I plug-in the x
$$ \begin{aligned} L_{\lambda_2} = x\Sigma x^T - \sigma^2 &= 0 \\ (\frac{1}{2\lambda_2})^2 (E(R) - \lambda_1 \vec{1}) \Sigma^{-1} \Sigma ((E(R) - \lambda_1 \vec{1}) \Sigma^{-1})^T &= \sigma^2\\ (\frac{1}{2\lambda_2})^2 (E(R) - \lambda_1 \vec{1}) \Sigma ( (E(R) - \lambda_1 \vec{1}))^T = \sigma^2 \end{aligned} $$
If I keep expanding the terms I cannot get a close form solution.
Note that I used the notations
$$ a = \vec{1} \Sigma^{-1} \vec{1}^T; \quad b = \langle E(R), \vec{1}\rangle_{\Sigma^{-1}} = E(R) \Sigma^{-1} \vec{1}^T $$