<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] The question posed was whether the OP's proof is correct.  Mark Bennet pointed out one flaw in comments, namely that you can't divide by $x-y$ if $x=y$.  It's worth noting another problem with the proof:  You can indeed suppose wlog that $x\ge y$, from which $x^2\cdot x\ge x^2\cdot y$ follows (because $x^2\ge0$), but the full inequality, ending in $\ge y^2\cdot y=y^3$, does not.  If, for example, $x=2$ and $y=-1$, we do not have $2^2\cdot(-1)\ge(-1)^2\cdot(-1)$.
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