Maximizing linear objective over spectraplex I have the following instance of a semidefinite program (SDP):
$$
d + f \to \max
$$
$$
a + b + c = 1
$$
$$
\begin{pmatrix}
a & d & e\\
d & b & f\\
e & f & c
\end{pmatrix} \succeq 0
$$
If I use Sylvester's criterion for this symmetric matrix, I get the following system:
$$
\begin{cases}
a, b, c \geq 0 \\
ab - d^2 \geq 0\\
bc - f^2 \geq 0\\
ac - e^2 \geq 0\\
abc - af^2 - be^2-cd^2 + 2def \geq 0
\end{cases}
$$
From the first three inequalites the following evaluation holds
$$
d \leq \sqrt{ab}, \qquad f \leq \sqrt{bc} \implies d + f \leq \sqrt{b}(\sqrt{a} + \sqrt{c})
$$
How can I maximize the right part of this inequality, or should I try to go another way to solve SDP?
I solved this problem in Mathematica and got that the following optimum

 A: Another way to solve the problem is to start, again, from the formulation
$$\begin{array}{ll} \underset{X}{\text{maximize}} & \langle A, X \rangle\\ \text{subject to} & \mbox{tr} (X) = 1\\ & {X} \succeq 0\end{array}$$
where
$${A} := \frac12 \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix}$$ used by Rodrigo de Azevedo.
The difference is that now, we consider the Lagrange multiplier $\mu$ and the dual problem
$$\min_{\mu\in\mathbb{R}}\max_{{X}\succeq0}\{\langle {A},{X}\rangle)+\mu(\langle I,{X}\rangle)-1).$$
We can see that strong duality holds in this case due to Slater's condition (i.e. there is an ${X}\succ0$ such that the problem is feasible) and that the primal is bounded.
The dual is then given by
$$-\mu^*:=\min_{\mu\in\mathbb{R}}-\mu\quad \mathrm{s.t.}\quad {A}+\mu I\preceq0$$
which is the same thing as saying that $-\mu^*$ is the maximum eigenvalue of the matrix ${A}$. By virtue of strong duality, the optimal value of the primal is also the maximum eigenvalue of ${A}$.
