lean_workbook_plus_39117

Prove that $\forall x,y,z>0$ we have: \n $\sum\frac{xy}{x^{2}+xy+y^{2}}\leq 1$ . \n \n here is what I found: \n We observe that $\frac{xy}{x^{2}+xy+y^{2}}\leq\frac{1}{3}$ which is equivalent to $(x-y)^{2}\geq 0$ you do the same with the others and you have the ineq. \n equality if $x = y = z$ . \n \n Do you have other solutions? \n \n cheers! \n \n By AM-GM,we have: \n $x^{2}+y^{2}\geq 2xy$ ,so $\frac{xy}{x^{2}+xy+y^{2}}\leq\frac{xy}{3xy}=\frac{1}{3}$ \n Similar we have $\sum\frac{xy}{x^{2}+xy+y^{2}}\leq 1$ .