I would like to show that $$A = x^T G(x) \tanh(x) \geq 0$$ (with equality only for $x=0$), for $x\in\mathbb{R}^n$. It is known that $G(x)\in\mathbb{R}^{n\times n}$ is a symmetric and positive definite matrix. Or come up with a counter-example if it does not hold.
I tried transforming to the associated quadratic forms $$A = \frac{1}{4}(x+\tanh(x))^T G(x) (x+\tanh(x)) - \frac{1}{4}(x-\tanh(x))^T G(x) (x-\tanh(x))$$
I thought since $x$ and $\tanh(x)$ have the same sign, it should be obvious to conclude, but I might be missing something. Is there a nice way to actually show it?
A brainless computer search for counterexamples yields this one:
Take
$$G=\begin{pmatrix}6 & 3 & -1 \\ 3 & 10 & 3 \\ -1 & 3 & 2\end{pmatrix}$$
and
$$x=\begin{pmatrix} -1 \\ 1 \\ -10\end{pmatrix}.$$
Then $x^T G \tanh x\simeq-1382$.