Lower bound on Hessian, mean-value theorem Let $x \mapsto f(x) \in \mathcal{C}^2$ be convex, i.e. $\forall x \in \mathbb{R}^n$, $\nabla^2f(x) \succeq 0$. Let $A \in \mathbb{R}^{m \times n}$ and suppose $\nabla^2f(x) + A^\top A \succ 0$. Is it true that
\begin{equation}
\int_{0}^{1} \nabla^2f(x^\ast + \tau(x - x^\ast))d\tau + A^\top A \succ 0
\end{equation}
for all $x,x^\ast \in \mathbb{R}^n$ and $\tau \in [0,1]$ (element-wise integration)? Does the first inequality imply the second one?

Yes, it is true.
Proof: Assume by contradiction that it is not true. Therefore, there exists $z, x, x^\ast \in \mathbb{R}^n$ such that:
$$
\int_{0}^{1} z^T\nabla^2f(x^\ast + \tau(x - x^\ast))zd\tau + z^TA^\top Az = 0
$$
Since both the terms are non-negative, they should both be zero. Therefore, we have concluded that for $y = x^\ast + \tau(x - x^\ast)$ we have $z^T(\nabla^2f(y) + A^TA)z = 0$ which contradicts with the assumption.
Note: In this proof I have used a concept from real analysis that if the integral of a non-negative function is zero, then the set on which the function is not equal to zero is of measure zero. You also need to add continuity of $\nabla^2f(x)$ to be able to conclude that $z^T\nabla^2f(x^\ast + \tau(x - x^\ast))z = 0$ which is the case here.
