For what points in $\mathbb R^2$ does the triangular condition fail for a 1/2-metric? It is known that for $p<1$, $d_p(x,y)=\big(\sum_{i=1}^n |y_i-x_i|^p\big)^{1/p}$ is not a metric. In the case of 2 dimensions and $p=1/2$ it seems rather hard to find a counterexample where the triangle condition fails, but there are such points. Can one obtain a general rule for which $x$,$y$,$z$ such that the following holds $d_{1/2}(x,z) > d_{1/2}(x,y)+d_{1/2}(y,z)$?
 A: I'm not sure what you mean by "a general rule", but you can get some intuition for this as follows. Let $T(A,B,C)$ be $d_{1/2}(A,B)+d_{1/2}(B,C)-d_{1/2}(A,C)$. Note that $T<0$ detects a failure of the triangle inequality.
Then you can easily plot the contours of the function $f_{P_0,P_1}(P)=\min(T(P_0,P_1,P),T(P_1,P_0,P))$ with fixed $P_0,P_1$; in particular, maybe you're interested in studying the properties of the curves $f_{P_0,P_1}(P)=0$ as $P_0$ and $P_1$ vary?
In the following figures (made in SageMath, code below), $P_0,P_1$ are joined by a green segment and the black contour line shows the locus of points where $f_{P_0,P_1}=0$. The regions where $f>0$ are colored reddish and those where $f<0$ are colored bluish (I couldn't be bothered to get the color scale perfectly right, sorry).
 

Code:
def disthalf(x,y):
    return (sqrt(abs(x[0]-y[0]))+sqrt(abs(x[1]-y[1])))^2

def triangletest(P0,P1,P):
    q1 = disthalf(P0,P1)+disthalf(P0,P)-disthalf(P,P1)
    q2 = disthalf(P0,P1)+disthalf(P1,P)-disthalf(P,P0)
    return min_symbolic(q1,q2)

x,y = var('x,y')
P0=(0,0)
P1=(1,1)

Cfill=contour_plot(triangletest(P0,P1,(x,y)),
             (x,min(P0[0],P1[0])-1,max(P0[0],P1[0])+1),
                   (y,min(P0[1],P1[1])-1,max(P0[1],P1[1])+1), 
                   cmap='coolwarm', fill=True, contours=10, plot_points=150)

C0=contour_plot(triangletest(P0,P1,(x,y)),
             (x,min(P0[0],P1[0])-1,max(P0[0],P1[0])+1),
                (y,min(P0[1],P1[1])-1,max(P0[1],P1[1])+1), 
                contours=[0],fill=False, plot_points=250)

C1 = line([P0,P1],color='green')

(Cfill+C0+C1).show(figsize=5)

A: Your metric is translation invariant, and so it suffices to find failures of the norm inequality $\|u+v\|_p \leq \|u\|_p + \|v\|_p$ where $\|u\|_p = d_p(u, 0)$. 
A simple example for $p = 1/2$ is $u = (1, 0), v = (0, 1)$, where $\|u + v\|_p = (1 + 1)^2 = 4 > 2 = \|u\|_p + \|v\|_p$. (Thus, take your triangle to be given by the points $x = (0, 0)$, $y = (0, 1)$, and $z = (1, 1)$.)
The following proposition is relevant and useful; my example is derived from the fact that the line segment (and particularly the midpoint) between $(0, 1)$ and $(1, 0)$ is outside of the "unit ball", which I hope you have a picture of. 
Proposition: If $V$ is a real or complex vector space and $N: V \to [0, \infty)$ is a function satisfying the scaling condition $N(\alpha \cdot v) = |\alpha| N(v)$ and the separability condition $N(v) = 0$ implies $v = 0$, then the following conditions are equivalent: 


*

*The norm inequality $N(x + y) \leq N(x) + N(y)$ is satisfied, 

*The unit ball $\{x \in V: N(x) \leq 1\}$ is convex, 

*If $N(u) = 1 = N(v)$, then $N(tu + (1-t)v) \leq 1$ whenever $0 \leq t \leq 1$. 
It's practically trivial to see that the first condition implies the second and the second implies the third, so let's show that the third implies the first (the implication most relevant to your question). 
Let's assume $x, y$ are both non-zero, else the norm inequality is trivially satisfied. By separability, $N(x), N(y)$ are non-zero, and so putting $u = \frac{x}{N(x)}$ and $v = \frac{y}{N(y)}$, we have $N(u) = 1 = N(v)$ by the scaling property. Then 
$$\frac{x+y}{N(x) + N(y)} = \frac{N(x)}{N(x) + N(y)} \frac{x}{N(x)} + \frac{N(y)}{N(x) + N(y)} \frac{y}{N(y)} = t u + (1-t)v$$ 
where $t = \frac{N(x)}{N(x) + N(y)} \in [0, 1]$. Assuming the third condition, we infer 
$$N\left(\frac{x+y}{N(x) + N(y)}\right) = N(tu + (1-t)v) \leq 1$$ 
whence by scaling, $\frac{N(x + y)}{N(x) + N(y)} \leq 1$, which gives the first condition. 
