1. **Stating the problem:** Solve the inequality $$\frac{1}{2}\left(x-\frac{2}{3}\right)^2 + x\left(x-\frac{2}{3}\right)\left(x+\frac{2}{3}\right) - x^3 \ge \frac{x}{2}\left(x-\frac{2}{3}\right) - \frac{8}{27}.$$ 2. **Rewrite and simplify each term:** - Expand the square term: $$\left(x-\frac{2}{3}\right)^2 = x^2 - 2 \cdot x \cdot \frac{2}{3} + \left(\frac{2}{3}\right)^2 = x^2 - \frac{4}{3}x + \frac{4}{9}.$$ - Multiply by $\frac{1}{2}$: $$\frac{1}{2}\left(x^2 - \frac{4}{3}x + \frac{4}{9}\right) = \frac{1}{2}x^2 - \frac{2}{3}x + \frac{2}{9}.$$ - Expand the product $x\left(x-\frac{2}{3}\right)\left(x+\frac{2}{3}\right)$: Note that $(x-\frac{2}{3})(x+\frac{2}{3}) = x^2 - \left(\frac{2}{3}\right)^2 = x^2 - \frac{4}{9}.$ So, $$x \cdot \left(x^2 - \frac{4}{9}\right) = x^3 - \frac{4}{9}x.$$ 3. **Rewrite the left side:** $$\frac{1}{2}x^2 - \frac{2}{3}x + \frac{2}{9} + x^3 - \frac{4}{9}x - x^3 = \frac{1}{2}x^2 - \frac{2}{3}x + \frac{2}{9} - \frac{4}{9}x.$$ Note that $x^3 - x^3 = 0$ cancels out. 4. **Simplify the $x$ terms:** $$- \frac{2}{3}x - \frac{4}{9}x = - \frac{6}{9}x - \frac{4}{9}x = - \frac{10}{9}x.$$ So the left side is: $$\frac{1}{2}x^2 - \frac{10}{9}x + \frac{2}{9}.$$ 5. **Rewrite the right side:** $$\frac{x}{2}\left(x - \frac{2}{3}\right) - \frac{8}{27} = \frac{x}{2}x - \frac{x}{2} \cdot \frac{2}{3} - \frac{8}{27} = \frac{1}{2}x^2 - \frac{1}{3}x - \frac{8}{27}.$$ 6. **Set up the inequality:** $$\frac{1}{2}x^2 - \frac{10}{9}x + \frac{2}{9} \ge \frac{1}{2}x^2 - \frac{1}{3}x - \frac{8}{27}.$$ 7. **Subtract $\frac{1}{2}x^2$ from both sides:** $$\cancel{\frac{1}{2}x^2} - \frac{10}{9}x + \frac{2}{9} \ge \cancel{\frac{1}{2}x^2} - \frac{1}{3}x - \frac{8}{27}$$ which simplifies to $$- \frac{10}{9}x + \frac{2}{9} \ge - \frac{1}{3}x - \frac{8}{27}.$$ 8. **Bring all terms to the left side:** $$- \frac{10}{9}x + \frac{2}{9} + \frac{1}{3}x + \frac{8}{27} \ge 0.$$ 9. **Combine like terms:** - Combine $x$ terms: $$- \frac{10}{9}x + \frac{1}{3}x = - \frac{10}{9}x + \frac{3}{9}x = - \frac{7}{9}x.$$ - Combine constants: $$\frac{2}{9} + \frac{8}{27} = \frac{6}{27} + \frac{8}{27} = \frac{14}{27}.$$ So inequality is: $$- \frac{7}{9}x + \frac{14}{27} \ge 0.$$ 10. **Isolate $x$:** $$- \frac{7}{9}x \ge - \frac{14}{27}.$$ 11. **Divide both sides by $-\frac{7}{9}$, remembering to flip the inequality sign because dividing by a negative:** $$x \le \frac{- \frac{14}{27}}{- \frac{7}{9}} = \frac{14}{27} \times \frac{9}{7} = \frac{14 \times 9}{27 \times 7} = \frac{126}{189} = \frac{2}{3}.$$ **Final answer:** $$\boxed{x \le \frac{2}{3}}.$$