Solve $x^2 + (y-1)^2 + (x-y)^2 - \frac{1}{3} = 0.$ I came across an interesting equation with two variables $x,y\in\mathbb{R}$, $x^2 + (y-1)^2 + (x-y)^2 - \frac{1}{3} = 0.$
This, once expanded, can be simplified to $3x^2 - 3xy + 3y^2 - 3y + 1=0.$
How can one proceed to solve it algebraically? The solution according to Wolfram is $(x,y)=(\frac{1}{3},\frac{2}{3}).$

You can use the Cauchy-Schwarz Inequality:
$$(1+1+1)\big(x^2+(1-y)^2+(y-x)^2\big)\geq \big(1\cdot x+1\cdot(1-y)+1\cdot(y-x)\big)^2=1^2=1\,.$$
Thus, 
$$x^2+(1-y)^2+(y-x)^2\geq\frac{1}{3}\,.$$
The inequality becomes an equality if and only if $$\frac{x}{1}=\frac{1-y}{1}=\frac{y-x}{1}\,,$$ or equivalently $$(x,y)=\left(\frac{1}{3},\frac{2}{3}\right)\,.$$
