<s>[INST] Inequality: $x^2+y^2+xy\ge 0$ I want to prove that $x^2+y^2+xy\ge 0$ for all $x,y\in \mathbb{R}$.
My "proof":  Suppose wlog that $x\ge y$, so $x^2\cdot x\ge x^2\cdot y\ge y^2\cdot y=y^3$ (because $x^2\ge 0$ so we can multiply both sides by it without changing the inequality sign) giving $x^3\ge y^3$. Substracting we have $x^3-y^3\ge 0$ or $(x-y)(x^2+xy+y^2)\ge 0$. Since $x\ge y$ we can divide by $x-y$ to get $x^2+xy+y^2\ge 0$.
Is it right?
Thanks for your help!
 [/INST] A homogeneous bivariate second-degree polynomial with a negative discriminant can be only everywhere non-negative or everywhere non-positive over $\mathbb{R}^2$. Since in $(x,y)=(1,1)$ the inequality holds as $\geq$, for every $(x,y)\in\mathbb{R}^2$ we have $x^2+xy+y^2\geq 0$.
</s>