It is very straightforward to show that $$a^3+b^3\geq \frac{1}{3}(a+b)(a^2+ab+b^2),$$since it is equivalent to $$3a^2-3ab+3b^2 \geq a^2+ab+b^2$$which is equivalent to $$a^2+b^2 \geq 2ab,$$which is obvious. Therefore, $$\frac{a^3+b^3}{a^2+ab+b^2} \geq \frac{a+b}{3}.$$Summing cyclicly, we get $$\sum_{cyc} \frac{x^3+y^3}{x^2+xy+y^2} \geq \sum_{cyc} \frac{x+y}{3} = \frac{2x+2y+2z}{3}.$$But by AM-GM $x+y+z \geq 3,$ so $$\sum_{cyc} \frac{x^3+y^3}{x^2+xy+y^2} \geq \frac{2x+2y+2z}{3} \geq 2. \Box$$edit: completed solution

parmenides51 wrote: Let $x, y, z$ be positive real numbers such that $xyz = 1$. Prove that: $$\frac{x^3 +y^3}{x^2 + xy + y^2}+\frac{ y^3+z^3}{y^2 + yz + z^2}+\frac{z^3 + x^3}{z^2 + zx + x^2}\ge 2.$$ Let $x, y, z$ be positive real numbers such that $xyz\ge1$. Prove that $$\frac{x^3 +y^3}{x^2 + xy + y^2}+\frac{ y^3+z^3}{y^2 + yz + z^2}+\frac{z^3 + x^3}{z^2 + zx + x^2}\ge 2$$(Romania 1997) Let $x, y, z$ be positive real numbers such that $xyz = 1$. Prove that $$\frac{x^{9}+y^{9}}{x^{6}+x^{3}y^{3}+y^{6}}+\frac{y^{9}+z^{9}}{y^{6}+y^{3}z^{3}+z^{6}}+\frac{z^{9}+x^{9}}{z^{6}+z^{3}x^{3}+x^{6}}\geq 2$$Let a,b,c>0. Prove that $$\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+ \frac{c^3}{c^2+ca+a^2} \geq \frac{a+b+c}{3}$$Let $a,b,c$ to be three non-negative real numbers such that $ a+b+c=3$ .Prove that :$$\frac{a^3+b^3}{a^2+\lambda ab+b^2}+\frac{b^3+c^3}{b^2+\lambda bc+c^2}+\frac{c^3+a^3}{c^2+\lambda ca+a^2} \ge \frac{6}{\lambda+2}.(\lambda >0)$$https://artofproblemsolving.com/community/c6h1371586p7559341