Inspired by Joel Tan

First look at cases where $x=y,y=z,x=z,$ or $x=y=z$ . The triangle $\Delta ABC$ is a line or a point and has an area of zero. This would be the minimum area so to get the maximum area $x\neq y \neq z$ . We then restrict $x,y$ and $z$ to the following: $0\leq x < y< z \leq 1.$

We can then graph points $A, B$ and $C$ on a graph with axes $u$ and $v$ with general points $(u,v)$ . $u$ being the horizontal axis and $v$ being the vertical axis.

Point $A$ is at $(x+y,x^2+y^2)$ , Point $B$ is at $(x+z,x^2+z^2)$ , and Point $C$ is at $(y+z,y^2+z^2)$ .

If we extend a horizontal line at $v=x^2+y^2$ . It intersects $u=x+z$ and $u=y+z$ at point $B^\prime$ and $C^\prime$ respectively.
If we extend a horizontal line at $v=x^2+z^2$ it intersects $u=y+z$ at Point $C^{\prime\prime}$ .

$B^\prime = (x+z,x^2+y^2)$

$C^\prime = (y+z,x^2+y^2)$

$C^{\prime\prime} = (y+z,x^2+z^2)$

$\Delta ABB^\prime, \Delta BCC^{\prime\prime}$ and $\Delta ACC^{\prime\prime}$ all form right triangles. $BB^{\prime} C^{\prime} C^{\prime\prime}$ forms a rectangle.

The area of $\Delta ABC$ is $| \Delta ACC^{\prime\prime} -\Delta ABB^\prime-\Delta BCC^{\prime\prime} - BB^{\prime} C^{\prime} C^{\prime\prime} |$ . It is absolute value since Point $B$ can lie above or below the line segment $AC$

$\Delta ACC^{\prime\prime} =\frac{(y^2+z^2-(x^2+y^2))(y+z-(x+y))}{2}=\frac{(z^2-x^2)(z-x)}{2}$

$\Delta ABB^\prime =\frac{(x^2+z^2-(x^2+y^2))(x+z-(x+y))}{2}=\frac{(z^2-y^2)(z-y)}{2}$

$\Delta BCC^{\prime\prime} =\frac{(y^2+z^2-(x^2+z^2))(y+z-(x+z))}{2}=\frac{(y^2-x^2)(y-x)}{2}$

$BB^{\prime} C^{\prime} C^{\prime\prime} = (x^2+z^2-(x^2+y^2))(y+z-(x+z))= (z^2-y^2)(y-x)$ .

We get rid of the absolute value sign and just look for most positive or negative value of $\Delta ACC^{\prime\prime} -\Delta ABB^\prime-\Delta BCC^{\prime\prime} - BB^{\prime} C^{\prime} C^{\prime\prime}$ .

= $\Delta ACC^{\prime\prime} -\Delta ABB^\prime-\Delta BCC^{\prime\prime} - BB^{\prime} C^{\prime} C^{\prime\prime}$

= $\frac{(z^2-x^2)(z-x)}{2} -\frac{(z^2-y^2)(z-y)}{2} -\frac{(y^2-x^2)(y-x)}{2}-(z^2-y^2)(y-x)$

= $\frac{(z^2-x^2)(z-x)}{2} -\frac{(z^2-y^2)(z-y)}{2} -\frac{(z^2-y^2)(y-x)}{2}-\frac{(y^2-x^2)(y-x)}{2}-\frac{(z^2-y^2)(y-x)}{2}$

Factoring yields

= $\frac{(z^2-x^2)(z-x)}{2} -\frac{(z^2-y^2)(z-y+y-x)}{2}-\frac{(y^2-x^2+z^2-y^2)(y-x)}{2}$

= $\frac{(z^2-x^2)(z-x)}{2} -\frac{(z^2-y^2)(z-x)}{2}-\frac{(z^2-x^2)(y-x)}{2}$

= $\frac{((z^2-x^2)-(z^2-y^2))(z-x)}{2}-\frac{(z^2-x^2)(y-x)}{2}$

= $\frac{(y^2-x^2)(z-x)}{2}-\frac{(z^2-x^2)(y-x)}{2}$

= $\frac{(y-x)(z-x)(y+x-(z+x))}{2}$

= $\frac{(y-x)(z-x)(y-z)}{2}$ .

So the area of $\Delta ABC$ is $\frac{(y-x)(z-x)(y-z)}{2}$ . First to maximize this you want to maximize $y-x,$ and $z-x$ . This can be done by making $x=0$ .

So the area of $\Delta ABC$ is now $\frac{yz(y-z)}{2}$ with $y+z=1$ . Substitute $z$ for $1-y$ .

$\Delta ABC = f(y)= \frac{y(1-y)(2y-1)}{2}$ for $0 < y <1$ .

The maximum of this is when the derivative of $f(y)$ equals zero.

$f^{\prime}(y)= -3y^2+3y-\frac{1}{2}=0$ at $y= \frac{3-\sqrt{3}}{6}$ or $\frac{3+\sqrt{3}}{6}$ .

$f( \frac{3-\sqrt{3}}{6}) = -\frac{1}{12\sqrt3}$ .

$f( \frac{3+\sqrt{3}}{6}) = \frac{1}{12\sqrt3}$ .

This means that the maximum area of $\Delta ABC$ is $\frac{1}{12\sqrt3}$ .

Thus $a=1,b=12,c=3$ and $a+b+c=\boxed{16}$

Consider the triangle with vertices $(x, x^2)$ , $(y,y^2)$ , and $(z, z^2)$ . Note that its medial triangle maps to $\triangle ABC$ through a homothety with a factor of $2$ w.r.t the origin, so our new triangle has the same area as $\triangle ABC$ , so we can just look at this new triangle.

By the Shoelace theorem, we see that the area is $[ABC]=\dfrac{1}{2}\left|\sum_{cyc}x^2y-\sum_{cyc}xy^2\right|$ which factors as $\dfrac{1}{2}|(x-y)(y-z)(x-z)|$ Now note that $y\ge x\ge z$ and $x\ge y\ge z$ are equivalent because of the absolute value, so WLOG $x\ge y\ge z$ which means the expression is positive and we just need to maximize $\dfrac{1}{2}(x-y)(y-z)(x-z)$ . Applying BW let $y=z+u$ and $x=z+u+v$ where $z,u,v$ are non-negative reals so our question is: given $3z+2u+v=1$ maximize $\dfrac{1}{2}uv(u+v)$ Clearly then we must have $z=0$ and thus $v=1-2u$ and we want to maximize $\dfrac{1}{2}u(1-2u)(1-u)$ given $0\le u\le \dfrac{1}{2}$ But taking a derivative gives that the maximum happens at $\dfrac{1}{2}u(1-2u)(1-u) =\boxed{ \dfrac{1}{12\sqrt{3}}}$ and we're done.