Proving homogenous quadratic inequalities

Question

Okay, I'm having trouble proving this: $$5x^2-4xy+6y^2 \ge 0, \text{ where } x,y \in \mathbb{R}$$ I have tried a few values of $x$ and $y$ and I find that is true. EX: $x=y=0$ which make the equation $= 0$

I have tried factoring it but i find that the factors will involve imaginary numbers.

So i would like a way to prove this inequality.

Answer 1

$$ 5x^2-4xy+6y^2 = 2(x-y)^2 + 3x^2 + 4y^2, $$ where each term is non-negative

Answer 2

If $y=0$ it's obvious. Let's assume $y \not = 0$:

$5x^2-4xy+6y^2 \gt 0 \iff 5(\frac{x}{y})^2-4\frac{x}{y}+6 \gt 0 \tag1$ Notation $\frac{x}{y} = t$ and (1) becomes:

$5t^2-4t+6 \gt 0 \tag2$

Note that expression $5t^2-4t+6$ has a negative discriminant.