Strong duality and KKT for SDP with inequality constraints The standard form of semidefinite program (SDP) is
\begin{align*}
p^* = \inf C \bullet X\\
s.t. A_i \bullet X = b_i\\
X \succeq 0
\end{align*}
where $C, A_i$ are symmetric matrices. The dual SDP is
\begin{align*}
d^* = \sup y^T b\\
s.t. C - \sum y_i A_i \succeq 0
\end{align*}
It's known (for example, see Ben-Tal & Nemirovski, Lectures on Modern Convex Optimization) that when there exists $X \succ 0$ strictly feasible for the primal, $C - \sum y_i A_i\succ 0$ strictly feasible for the dual, then we have strong duality: optimizers for both problems exist, and $p^* = d^*$. Moreover, a primal dual pair is optimal if and only if it satisfies the KKT condition $X \bullet (C - \sum y_i A_i) = 0$.
My question: do the same results hold for SDP with inequality constraints (i.e., $A_i \bullet X = b_i$ replaced by $A_i \bullet X \leq b_i$)? Specifically, consider the primal
\begin{align*}
p^* = \inf C \bullet X\\
s.t. A_i \bullet X \leq b_i\\
X \succeq 0
\end{align*}
and dual
\begin{align*}
d^* = \sup y^T b\\
s.t. C - \sum y_i A_i \succeq 0\\
y_i \leq 0
\end{align*}
Assume there exists $X \succ 0$ feasible for the primal, and $y_i \leq 0$ such that $C - \sum y_i A_i \succ 0$. Is it true that

*

*optimal solutions for both the primal and dual are achieved

*$p^* = d^*$

*a primal dual pair $(X, y_i)$ is optimal iff $X \bullet (C - \sum y_i A_i) = 0$?

All of the literature I've seen only discusses strong duality for SDP with equality constraints, so I can't find an answer for my question.

Below are my thoughts: let $X' = \begin{pmatrix}
X & 0\\
0 & D
\end{pmatrix}$, $A_i' = \begin{pmatrix}
A_i & 0\\
0 & V_i
\end{pmatrix}$, $C' = \begin{pmatrix}
C & 0\\
0 & 0
\end{pmatrix}$, where $D$ is diagonal, $V_i$ has a $1$ at $ii$-entry, zero everywhere else. Then the primal can be reformulated as
\begin{align*}
p' = \inf C' \bullet X'\\
s.t. A_i' \bullet X' = b_i\\
X' \succeq 0
\end{align*}
and the dual of this new problem is
\begin{align*}
d' = \sup y^T b\\
s.t. C' - \sum y_i A_i' \succeq 0
\end{align*}
Since $C' - \sum y_i A_i' = \begin{pmatrix}
C - \sum y_i A_i & 0\\
0 & diag(-y_i)
\end{pmatrix} \succeq 0$, we know that $y_i \leq 0$. So this is just the dual of the original SDP.
For strong duality, suppose there exists $X \succ 0$ feasible for the original problem. I'd like to show that there exists $X' \succ 0$ feasible for the new primal. We can take $X' = \begin{pmatrix}
X & 0\\
0 & D
\end{pmatrix}$, with $D$ containing strictly positive diagonal entries. However, this is not going to work unless $A_i \bullet X < b_i$ for all $i$. This is one place where I am stuck.
Another place I'm unsure is the KKT condition. Assuming for now that the above assumption holds. Is it still true that $(X, y_i)$ optimal iff $X \bullet (C - \sum y_i A_i) = 0$? Here's my attempt.
First assume that $(X, y_i)$ is an optimal primal dual pair. Put $d_i = b_i - X \bullet A_i$, $X' = \begin{pmatrix}
X & 0\\
0 & diag(d_i)
\end{pmatrix}$. Then $(X', y_i)$ is an optimal primal dual pair for $p'$ and $d'$. We have
\begin{align*}
0 & = X' \bullet (C' - \sum y_i A_i')\\
& = X \bullet (C - \sum y_i A_i) - \sum y_i d_i
\end{align*}
since $X \succeq 0$, $C - \sum y_i A_i \succeq 0$, $d_i \geq 0$, $y_i \leq 0$, we must have $X \bullet (C - \sum y_i A_i) = 0$ and $\sum y_i d_i = 0$.
Conversely, assume that $X \bullet (C - \sum y_i A_i) = 0$. Put $d_i = b_i - X \bullet A_i$, $X' = \begin{pmatrix}
X & 0\\
0 & diag(d_i)
\end{pmatrix}$. Then
\begin{align*}
X' \bullet (C' - \sum y_i A_i') & = X\bullet( C - \sum y_i  A_i) - \sum y_i d_i\\
& = - \sum y_i d_i
\end{align*}
Again, I am stuck at this step.
 A: Brian Borchers' comment addresses the first question, pointing out that Slater's condition requires strict feasibility in all inequalities.
For the second question, the KKT conditions are more than just $X \bullet (C - \sum y_i A_i) = 0$ (stationarity) alone. Wikipedia additionally lists (adapting the general conditions to the case of SDPs, with your notation and sign conventions)

*

*Primal feasibility: $d_i(X) = b_i - A_i \bullet X \geq 0$, $X \succeq 0$

*Dual feasibility: $y_i \leq 0$, $C - \sum y_i A_i \succeq 0$

*Complementary slackness: $\sum_i y_i d_i(X) = 0$
These, together with stationarity and a constraint qualification (such as Slater's condition), are jointly necessary conditions fulfilled by all optima.
Your proof of $(X, y_i)\text{ optimal }\Rightarrow \mathrm{KKT}(X, y_i)$ (KKT necessary) shows how optimality implies primal and dual feasibility at $X$, and uses it to prove complementary slackness, and then stationarity.
For $\mathrm{KKT}(X, y_i) \Rightarrow (X, y_i)\text{ optimal}$ (KKT sufficient), as you have noted, you need complementary slackness to prove the stationarity of $(X', y_i')$ from the stationarity of $(X, y_i)$. Since stationarity of $(X', y_i')$ alone is sufficient for its equality-constrained problem, whereas inequality-constrained problems require all KKT conditions to be fulfilled, it is not surprising that fulfilling some of the KKT conditions for $(X, y_i)$ does not imply fulfilling the condition for $(X', y_i')$.
As a side note, even for problems with only equality constraints, stationarity and primal feasibility are jointly necessary.
