Let $A\in \mathbb{R}^{n\times n}$ be a positive definite matrix. Then find
\begin{equation} \max_{|x_i|\leq1}x^TAx \end{equation}
Here I want to use KKT conditions to show that $|x_i|=1$ is an optimal solution.
Derivative of objective function $\nabla f(x^*)=2Ax$ and derivative of inequality constraint function is $\sum_{i=1}^{n}\mu_i\nabla g_i(x^*)=[\pm\mu_1 \quad \pm\mu_2 \quad \cdots \quad \pm\mu_n]'$, so KKT conditions are
\begin{equation} 2Ax-[\pm\mu_1 \quad \pm\mu_2 \quad \cdots \quad \pm\mu_n]'=0; \end{equation} \begin{equation} \mu_i\geq0; \end{equation} \begin{equation} \mu_ig_i(x^*)=0; \end{equation} So if $2Ax$ is non-zero element-wise, then $\mu_i \neq0$, thus $g_i(x^*)=0\quad$ i.e. $|x_i|=1$ is optimal solution.
But how to show that $2Ax$ is indeed non-zero element-wise?