Let $A$ and $B$ be two symmetric matrices from $\mathbb{R}^{n \times n}$ such that $\operatorname{tr}{A} > 0$ and $\operatorname{tr}B < 0$.
How one can prove that there exists a vector $x \in \mathbb{R}^n$ such that $(Ax, x) > 0$ and $(Bx, x) < 0$?
Since the matrices are real and symmetric we can select a basis in which $A$ is diagonal. In this case, its trace doesn't change. If $A = \operatorname{diag}(d_1, d_2, \dots, d_n)$, then we can note that $$(Ax, x) = \sum\limits_{i=1}^n d_ix_i^2$$ So, it is easy to make this one positive by chosing appropriate $x$, say, $x = (1, 1, \dots, 1)^T$. But we also need to ensure that $(Bx, x) < 0$. So the question now is how to choose $x$ so that both inequalities hold.
For those who are interested, there turns out to be a pretty elegant solution to this problem that involves probabilistic methods.
As were said in the post, we can choose a basis in which $A$ is diagonal. Let $X$ be a random vector: its i-th coordinate $X_i$ equals to 1 with probability $\dfrac{1}{2}$ and $-1$ with probability $\dfrac{1}{2}$ (all coordinates are independent). Then,
$$(Ax, x) = \sum\limits_{i=1}^n d_ix_i^2 = \sum\limits_{i=1}^n d_i = \operatorname{tr}A > 0$$
But
$$\operatorname{\mathbb{E}}(Bx, x) = \operatorname{\mathbb{E}}\left(\sum\limits_{ij} B_{ij}x_ix_j\right) = \sum\limits_{ij} B_{ij}\operatorname{\mathbb{E}}(x_ix_j) = \sum\limits_{i} B_{ii}\operatorname{E}(x_i^2) = \operatorname{tr}(B) < 0$$
It means that there exists such a realization of $x$ that $(Bx, x) < 0$ (if it isn't true, the expectation would be non-negative).