How to prove what values of $\beta$ would $f{\left (x,y \right )} = -\beta x y + x^{2} + y^{2}$ is always greater than 0 $\beta \in R$
$f{\left (x,y \right )} = -\beta x y + x^{2} + y^{2}$
For what values is $\beta$, $f(x,y) \geq 0$
Note: I know that when $-2 \leq \beta \leq 2$, $f(x,y) \geq 0$, but how do I prove it?
 A: Complete the square on $x$:
$$ x^2 - \beta xy + y^2 = (x-\beta y/2)^2 +\frac{4-\beta^2}{4}y^2. $$
The bracket is always positive, the $y^2$ second term can be made as large as you like compared to the first term, so for the sum to definitely be nonnegative, you need $4-\beta^2 \geq 0$, i.e. $-2\leq\beta\leq2$.
A: I'll give you a short sketch of the proof below. You might fill in the exact details if you like.
Assume $2>\beta>0$ ($\beta=0$ should not be too hard, and the other case works similarly, but you should check it though). Then, note that $-\beta x y \geq 0$ if $x$ and $y$ have different signs. Hence, you could claim that it is enough to prove the result for $x,y>0$. So we have now reduced the problem to something easier.
The next thing to do is to prove that there is a non-negative lower bound for your function, which then proves that your function is non-negative. I'll give you a hint here: look for a quadratic polynomial.
I hope this helps!
A: The equalities $$\cases{f(x,y)=(x-y)^2+(2-\beta)xy\\f(x,y)=(x+y)^2+(-2-\beta)xy\\}$$
prove respectively that $$\cases{\text{when $\beta>2$, any $x=y\neq0$ will make $f$ negative}\\ \text{when $\beta<2$, any $x=-y\neq0$ will make $f$ negative}}$$
Finally, when $-2\leq\beta\leq 2$, you have $f(x,y)$ comprised between $(x-y)^2$ and $(x+y)^2$ (no matter which of the two is greater).
A: You must have the following:
$$f(x,y) \ge 0 \Rightarrow -βxy+x^2+y^2 \ge 0$$
Take the discriminant with regard to x:
$$D=β^2y^2-4y^2=(β^2-4)y^2$$
In order for the above equation to be always positive the following should stand:
$$D \le 0 \Rightarrow (β^2-4)y^2 \le 0 \Rightarrow    β^2-4\le 0 \Rightarrow β^2\le 4\Rightarrow -2\le β\le 2 $$
