Suppose $f:\mathbb{R^2_+} \to \mathbb{R}$ is a function such that $f(1,1) = 1$, $f(x,y) = f(y,x)$ for all $(x,y) \in \mathbb{R^2_+}$ and $f$ is homogenous of degree $1$. Does there exist any such $f$ that also satisfies $f(p,q) > (p+q)$ for some point $(p,q) \in \mathbb{R^2_+}$?
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Since you don't require that the function be continuous, I think the answer is yes. For example, define $f(x, y)$ to be $2x$ on the line $y = \frac{x}{2}$, $x$ on the line $y = x$, $2y$ on the line $y = 2x$, and 0 everywhere else. Then for example at $(1, \frac{1}{2})$, $f(x, y) = 2$ while $x + y = \frac{3}{2}$.
You can also view this type of function using polar coordinates.
$$ f(r \cos \theta, r \sin \theta) = r g(\theta) $$
The homogeneity requirement is satisfied by the fact that $\frac{f}{r}$ does not depend on $r$. The requirement $f(1,1)=1$ forces that $g\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$. Otherwise, $g$ can be any function on the interval $\left(0,\frac{\pi}{2}\right)$. If $g$ is continuous, so is $f$. If $g$ is differentiable and your definition of $\mathbb{R}_{+}$ does not include $0$, then $f$ is also differentiable. And if at any value $0<\theta<\frac{\pi}{2}$ we have $g(\theta) > \cos \theta + \sin \theta$, this means that $f(x,y) < x+y$ using the usual relations $x=r\cos\theta$ and $y=r\sin\theta$.
If $(0,0)$ is in the domain, I think the only possible differentiable $f$ are plane equations $f(x,y) = ax+by$. This can still lead to values where $f(x,y) > x+y$. For example, $f(x,y) = 3x-2y$ gives $f(1,1)=1$ and $f(3,1)=7 > 3+1$.