1. **State the problem:** Solve the inequality $$x\left(\frac{1}{2}-\frac{1}{3}\right)^{-1} + \frac{(x-1)^2}{2} + \frac{1}{2}x^2 \geq 5x\left(\frac{1}{2}+\frac{1}{3}\right)^{-1} + \frac{(2x-1)^2}{4}.$$\n\n2. **Simplify the fractions inside the parentheses:**\nCalculate $$\frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}.$$\nCalculate $$\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}.$$\n\n3. **Rewrite the expression using these values:**\n$$x\left(\frac{1}{6}\right)^{-1} + \frac{(x-1)^2}{2} + \frac{1}{2}x^2 \geq 5x\left(\frac{5}{6}\right)^{-1} + \frac{(2x-1)^2}{4}.$$\n\n4. **Calculate the inverses:**\n$$\left(\frac{1}{6}\right)^{-1} = 6,$$\n$$\left(\frac{5}{6}\right)^{-1} = \frac{6}{5}.$$\n\n5. **Substitute back:**\n$$6x + \frac{(x-1)^2}{2} + \frac{1}{2}x^2 \geq 5x \cdot \frac{6}{5} + \frac{(2x-1)^2}{4}.$$\nSimplify right side term: $$5x \cdot \frac{6}{5} = 6x.$$\n\n6. **Rewrite inequality:**\n$$6x + \frac{(x-1)^2}{2} + \frac{1}{2}x^2 \geq 6x + \frac{(2x-1)^2}{4}.$$\n\n7. **Subtract $6x$ from both sides:**\n$$6x + \frac{(x-1)^2}{2} + \frac{1}{2}x^2 - 6x \geq 6x + \frac{(2x-1)^2}{4} - 6x,$$\nwhich simplifies to\n$$\frac{(x-1)^2}{2} + \frac{1}{2}x^2 \geq \frac{(2x-1)^2}{4}.$$\n\n8. **Multiply both sides by 4 to clear denominators:**\n$$4 \cdot \left(\frac{(x-1)^2}{2} + \frac{1}{2}x^2\right) \geq 4 \cdot \frac{(2x-1)^2}{4},$$\nwhich gives\n$$2(x-1)^2 + 2x^2 \geq (2x-1)^2.$$\n\n9. **Expand all squares:**\n$$(x-1)^2 = x^2 - 2x + 1,$$\n$$(2x-1)^2 = 4x^2 - 4x + 1.$$\n\n10. **Substitute expansions:**\n$$2(x^2 - 2x + 1) + 2x^2 \geq 4x^2 - 4x + 1.$$\n\n11. **Distribute:**\n$$2x^2 - 4x + 2 + 2x^2 \geq 4x^2 - 4x + 1.$$\n\n12. **Combine like terms on left:**\n$$4x^2 - 4x + 2 \geq 4x^2 - 4x + 1.$$\n\n13. **Subtract $4x^2 - 4x$ from both sides:**\n$$4x^2 - 4x + 2 - (4x^2 - 4x) \geq 4x^2 - 4x + 1 - (4x^2 - 4x),$$\nwhich simplifies to\n$$2 \geq 1.$$\n\n14. **Interpretation:**\nSince $$2 \geq 1$$ is always true, the inequality holds for all real values of $x$.\n\n**Final answer:**\n$$\boxed{\text{The inequality is true for all real } x.}$$