Prove that $x^2 + 5xy+7y^2 \ge 0$ for all $x,y \in\mathbb{R}$ This is probably really easy for all of you, but my question is how do I prove that $x^2 + 5xy+7y^2 \ge 0$ for all $x,y\in\mathbb{R}$
Thanks for the help!

You can’t prove that it’s always greater than $0$, because when $x=y=0$ it is $0$, but you can prove that it’s always greater than or equal to $0$ by completing the square:
$$x^2+5xy+7y^2=\left(x+\frac{5y}2\right)^2+\frac34y^2\;.$$
Since square are always at least $0$, so is their sum.
