$\sum(F\circ F)\geq x_1^2+\cdots+x_n^2$

Question

Let $A,B$ be $n\times n$ matrix, denote by $A\circ B=(a_{ij}b_{ij})$.

Let $C$ be an invertible real matrix. $D=diag(x_1,\cdots,x_n)$, $F=CDC^{-1}$. Show that $\sum(F\circ F)\geq x_1^2+\cdots+x_n^2$, where $\sum$ means the sum of all entries of $F\circ F$.

If $C$ is the identity matrix, it is OK. What about other matrices?

Answer 1

$ \def\L{\left}\def\R{\right}\def\LR#1{\L(#1\R)} \def\Diag#1{\operatorname{Diag}\LR{#1}} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\m#1{\left[\begin{array}{r}#1\end{array}\right]} $Here is a small $(n=3)\,$ counter-example $$\eqalign{ &C = \m{ -1 & 6 & -6 \\ 4 & 3 & 1 \\ -3 & 2 & 4 \\} \quad x = \m{1\\2\\6} \\ &F = C\cdot\Diag{X}\cdot C^{-1} = \frac 1{113}\m{ 425 & -174 & -336 \\ -14 & 186 & 102 \\ -151 & 182 & 406 \\} \\\\ &\trace{F\circ F} = \frac{425^2+186^2+406^2}{113^2} \approx 29.76404 \\ &\|x\|^2 = 41 \\ }$$