1. **State the problem:** Show that for all real numbers $x, y, z$, the inequality $$x^2 + y^2 + z^2 \geq xy + yz + zx$$ holds. 2. **Rewrite the inequality:** We want to prove: $$x^2 + y^2 + z^2 - (xy + yz + zx) \geq 0$$ 3. **Use algebraic manipulation:** Group terms to express the left-hand side as a sum of squares. Consider: $$\frac{1}{2} \big[(x - y)^2 + (y - z)^2 + (z - x)^2\big]$$ 4. **Expand the right side:** $$= \frac{1}{2} [(x^2 - 2xy + y^2) + (y^2 - 2yz + z^2) + (z^2 - 2zx + x^2)]$$ $$= \frac{1}{2} [2x^2 + 2y^2 + 2z^2 - 2(xy + yz + zx)]$$ $$= x^2 + y^2 + z^2 - (xy + yz + zx)$$ 5. **Conclude the inequality:** Since squares are always non-negative, $$(x - y)^2 \geq 0, \quad (y - z)^2 \geq 0, \quad (z - x)^2 \geq 0$$ Therefore, $$x^2 + y^2 + z^2 - (xy + yz + zx) = \frac{1}{2} \big[(x - y)^2 + (y - z)^2 + (z - x)^2\big] \geq 0$$ which proves the original inequality. **Final answer:** The inequality $$x^2 + y^2 + z^2 \geq xy + yz + zx$$ holds for all real numbers $x, y, z$.