How high can it go?

By completing squares, the given equation can be written as

$x^{2} - \dfrac{x}{2} + y^{2} - \dfrac{y}{2} = 0 \Longrightarrow \left(x - \dfrac{1}{4}\right)^{2} + \left(y - \dfrac{1}{4}\right)^{2} = \dfrac{1}{8}$ ,

which describes a circle with center $\left(\dfrac{1}{4}, \dfrac{1}{4}\right)$ and radius $\dfrac{1}{\sqrt{8}} = \dfrac{1}{2\sqrt{2}}$ .

We can then let $x - \dfrac{1}{4} = \dfrac{1}{2\sqrt{2}}\cos(\theta)$ and $y - \dfrac{1}{4} = \dfrac{1}{2\sqrt{2}}\sin(\theta)$ .

From this we have that

$x - y = \dfrac{1}{2\sqrt{2}}(\cos(\theta) - \sin(\theta)) = \dfrac{1}{2}\left(\dfrac{1}{\sqrt{2}}\cos(\theta) - \dfrac{1}{\sqrt{2}}\sin(\theta)\right) = \dfrac{1}{2}\sin\left(\dfrac{\pi}{4} - \theta \right)$ .

As the maximum of the sine function is $1$ , we see that the maximum of $x - y$ is $\dfrac{1}{2} = \boxed{0.5}$ .

$\begin{aligned}x+y&=2(x^2+y^2)\hspace{10mm}\small\color{#3D99F6}\text{Given}\\ x+y-(x+y)^2&=2(x^2+y^2)-(x^2+2xy+y^2)\\ &=x^2-2xy+y^2\\ &=(x-y)^2\\ \\\text{Let}(x+y)=t,\\ (x-y)^2 &=f(t)=t-t^2\\ f'(t)&=1-2t\\ f'(t)&=0 \implies t=\dfrac {1}{2}\\ f''(t)&=-2<0\\ \implies f(\dfrac {1}{2})=\dfrac{1}{4}& \text{ is a maxima}\\ (x-y)^2&\leq\dfrac{1}{4}\\ -\dfrac {1}{2}\leq x-y&\leq\dfrac{1}{2}\\ \text{Maximum}&\text{ value of }x-y=\boxed{\dfrac{1}{2}} \end{aligned}$

$2(x^2+y^2)=x+y \\ => 2(x+y)^2 -4xy=x+y \\ => 4xy=(x+y)- 2(x+y)^2$

$(x-y)^2=(x+y)^2-4xy=(x+y)-(x+y^2)$

Using the Completing-Square Rule-

$(x-y)^2=\dfrac{1}{4}-(x+y-1/2)^2$

Maximum value occurs at $x+y=1/2$ and maximum value of $x-y=\dfrac{1}{2}$

$2(x^2 + y^2) = x + y$

$\Rightarrow \left(x - \dfrac{1}{4}\right)^2 + \left(y - \dfrac{1}{4}\right)^2 = \dfrac{1}{8} = \left(x - \dfrac{1}{4}\right)^2 + \left(\dfrac{1}{4} - y\right)^2$

By Titu's Lemma ,

$\dfrac{1}{8} = \left(x - \dfrac{1}{4}\right)^2 + \left(\dfrac{1}{4} - y\right)^2 = \dfrac{\left(x - \dfrac{1}{4}\right)^2}{1} + \dfrac{\left(\dfrac{1}{4} - y\right)^2}{1} \geq \dfrac{(x - y)^2}{2}$

$\Rightarrow (x - y) \leq \dfrac{1}{2}$

With equality if $\left(x - \dfrac14\right) = \left(\dfrac14 - y\right) \Rightarrow \boxed{x + y = \dfrac12}$

$\begin{aligned} 2(x^2+y^2) &= x+y \\ 2((x+y)^{2}-2xy) &= (x+y) \\ 2(x+y)^{2}-4xy &= (x+y) \\ (x+y)^{2}-4xy &= (x+y)-(x+y)^{2} \\ (x-y)^{2} &= (x+y)-(x+y)^{2} \\ (x-y) &= \sqrt{ (x+y)-(x+y)^{2} } \end{aligned}$

$LHS$ maximum when $RHS$ maximum.

$RHS$ maximum when $(x+y) = - \dfrac{1}{ 2(-1) } = \dfrac{1}{2}$

Putting $(x+y) =\dfrac{1}{2}$ , we have $(x-y) = \sqrt{ \dfrac{1}{2}- \dfrac{1}{2}^{2} } = \sqrt{ \dfrac{1}{4} } = \dfrac{1}{2} = \boxed{0.5}$

From am-gm we get $x^{2}+y^{2}=\frac{1}{2}$ and $x+y=1$ , so we get maximum value of $x-y=\frac{1}{2}$