<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] One (simple) way:
let t = x2 + y2 + xy
then t - xy ≥ 0 since it is the sum of two real squares x2 + y2
and t + xy ≥ 0 since it is the square of the real (x + y) [since (x + y)2 = x2 + y2 + 2xy]
adding these, we get 2t ≥ 0, therefore t ≥ 0
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