Application of Jensen's Inequality-Positive Definite Matrices-Probability I'm given the question for $A$ positive definite matrix in $\mathbb{R}^n$, use Jensen's inequality to prove $(x^TAx)(x^TA^{-1}x)\geq1$ for unit vectors $x\in\mathbb{R}^n$. The hint is to think about $A$ diagonal and then diagonalize $A$ in the general case. (If $A$ is diagonal, $(x^TAx)(x^TA^{-1}x)=1$, but positive definite does not imply diagonalizable).

I know $x^TAx&gt;0$ for all $x\in\mathbb{R}^n$. But in order to use Jensen's inequality, I need to have a convex function. Here is what I have thought about:

Let $g(y)=1/y$ is a convex function, then $g(\mathbb{E}y)\leq\mathbb{E}(g(y))$. I want then to get $y=(x^TAx)(x^TA^{-1}x)$ to see $\frac{1}{(x^TAx)(x^TA^{-1}x)}\leq 1$. I know this does not work, but it is how I am trying to use Jensen.
Hint: If $D$ is diagonal with elements $d_1,d_2\ldots, d_n$, then $D^{-1}$ is diagonal with elements $\frac1{d_1},\ldots,\frac1{d_n}$. Then
$$x^TDx = \sum_i d_ix_i^2 = E(Y)$$
where $Y$ takes value $d_i$ with probability $x_i^2$. (Note that $x$ has norm 1, so the probabilities add up to 1.) What is the corresponding statement for $x^TD^{-1}x$? Now apply Jensen's inequality $Ef(Y)\ge f(E(Y))$ with a suitable choice of $f$. Spoiler below:


   $f(y)=\frac1y$.
Fix $x$, the map $B\xrightarrow{\quad\pi\quad}\left\langle x,Bx\right\rangle $
is positive definite on $M\left(n,\mathbb{C}\right)$. So by Cauchy-Schwarz inequality, 

\begin{equation}
\left|\pi\left(A^{1/2}A^{-1/2}\right)\right|^{2}\leq\pi\left(A^{1/2}A^{1/2}\right)\pi\left(A^{-1/2}A^{-1/2}\right),
\end{equation} i.e.,  $\left\langle x,Ax\right\rangle \left\langle x,A^{-1}x\right\rangle \geq1$, with $\left\Vert x\right\Vert =1$.