Consider $g_1$ and $g_2$ two Riemannian metrics on a differentiable manifold $M$ of dimension **$n\ge 4$**. Suppose locally $g_i=f_i\sum_{j=1}^ndx_j^2$, where $f_i:M\rightarrow \mathbb{R}$ are non negative functions. Suppose that both $g_1$ and $g_2$ have non positive sectional curvature.
Define the metric $\hat g$ on $M$ as $\hat g:=max\{f_1,f_2\}\sum_{j=1}^ndx_j^2$.

**Short question:** Will $(M,\hat g)$ have non positive curvature?

**More elaborate question:** A priori $\hat g$ has just continuous coefficients: we can ask if its intrinsic induced metric $d_{\hat g}$ has non positive Alexandrov curvature (it means that every $p\in M$ has an open neighborhood $U$ such that for every $x,y,z\in U$ $d_{\hat g}(z,m)^2\le \frac 1 2(d_{\hat{g}}(z,x)^2+d_{\hat{g}}(z,y)^2)-\frac 1 4d_{\hat{g}}(x,y)^2$, where $m$ is the midpoint between $x$ and $y$, definition taken from http://people.mpim-bonn.mpg.de/hwbllmnn/archiv/NPC0606.pdf)

Clearly in the open sets $U_1:=\{p\in M|f_1(p)>f_2(p)\}$ or $U_2:=\{p\in M|f_2(p)>f_1(p)\}$ this is true, because the metrics $g_1$ and $g_2$ have negative sectional curvature. But I don't know what happens in points where $f_1=f_2$.

Say $p\in M$ is such that $f_1(p)=f_2(p)$. Then every open neighborhood $U$ of $p$ will intersect both $U_1$ and $U_2$ and evaluating directly if the preceding inequality is true will be quite difficult. Is there a more straightforward way (than direct computation) to conclude that $d_{\hat{g}}$ has non positive Alexandrov curvature?