Let $X\in \mathbb R^d$ be a random vector and $\mu=E(X)\in \mathbb R^d$ be its mean. Let $\Sigma=E[(X-\mu)(X-\mu)^T]\in\mathbb R^{d\times d}$ be the variance-covariance matrix of X. Show that $\Sigma$ is positive semidefinite using Jensen's inequality.
Hint Show that $f(X)=a^TXX^Ta, a\in \mathbb R^d$ is convex.
Progress
I want to use the hint and the fact that the hessian of a convex function is positive semidefinite.
$$\begin{split}f(X)&=(a^TX)^2\\ \nabla f(x)&=2a^TXa\\ \nabla^2 f(x)&=2aa^T\end{split}$$
Is the last step justified? I sort of made it up, but it might be right (my reasoning: the derivative of $a^TX$ is a, and then you transpose the other a). Certrainly the dimension of it is right. I don't know what the derivative of $a^TXa$ wrt X would be.
$\nabla f(x)=2\left(\begin{matrix}a^TXa_1\\...\\a^TXa_d\end{matrix}\right)=2\left(\begin{matrix}(a_1X_1+...+a_dX_d)a_1\\...\\(a_1X_1+...+a_dX_d)a_d\end{matrix}\right)$
I always get confused about taking the derivative of this wrt X. Do I take wrt $x_1$ and put that in the first column or put that in the first row? (I think it doesn't matter since it's symmetric, but I would still like to know the answer for some reason.)
$\nabla^2 f(x)=2\left(\begin{matrix}a_1^2&a_1a_2&...&a_1a_d\\...\\a_da_1&a_da_2&...&a_d^2\end{matrix}\right)$
I think this equals $aa^T$ so we're good. (An outer product of a vector wrt itself is psd.) Now how does this help the original problem?