Maximum value of $|x-y|$ Given $x,y\in\mathbb R$ such that
$$5x^2+5y^2-6xy\ =\ 8$$
find the maximum value of $|x-y|$.
My attempt
$5x^2 - 6yx + (5y^2-8)\ =\ 0$
$x\ =\ \dfrac{6y\pm\sqrt{(6y)^2-4(5)(5y^2-8)}}{10} = \dfrac{6y\pm\sqrt{160-64y^2}}{10} = \dfrac{3y\pm2\sqrt{10-4y^2}}{10}$
$5y^2 - 6xy + (5x^2-8)\ =\ 0$
$y\ =\ \dfrac{3x\pm2\sqrt{10-4x^2}}{10}$
$|x-y|\ =\ \dfrac{\left|3(y-x)\pm2\sqrt{10-4y^2}\mp2\sqrt{10-4x^2}\right|}{10}$
What do I do from here?

We'll prove that $|x-y|\leq\sqrt2$.
Indeed, we need to prove that 
$$(x-y)^2\leq2\cdot\frac{5x^2+5y^2-6xy}{8}$$ or
$$(x+y)^2\geq0.$$
The equality occurs for $x+y=0$ and $x-y=\sqrt2$ for example.
Id est, the answer is $\sqrt2$.
