<s>[INST] 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 $(x,y)=(\frac{1}{3},\frac{2}{3}).$
 [/INST] $$3x^2+3(y-1)^2+3(x-y)^2-1=(x+y-1)^2+(2x-y)^2+(x-2y+1)^2\geq0.$$
The equality occurs for
$$x+y-1=2x-y=x-2y+1=0,$$ which gives $x=\frac{1}{3}$ and $y=\frac{2}{3}$.
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