1. **Problem:** Write any one relation that satisfies AM, GM, and HM between any two unequal positive numbers. 2. **Formula and Explanation:** For two positive numbers $a$ and $b$, the Arithmetic Mean (AM), Geometric Mean (GM), and Harmonic Mean (HM) satisfy the inequality: $$\text{HM} \leq \text{GM} \leq \text{AM}$$ where $$\text{AM} = \frac{a+b}{2}, \quad \text{GM} = \sqrt{ab}, \quad \text{HM} = \frac{2}{\frac{1}{a} + \frac{1}{b}}$$ This means the harmonic mean is always less than or equal to the geometric mean, which is always less than or equal to the arithmetic mean. 3. **Intermediate Work:** - Calculate AM: $\frac{a+b}{2}$ - Calculate GM: $\sqrt{ab}$ - Calculate HM: $\frac{2}{\frac{1}{a} + \frac{1}{b}} = \frac{2ab}{a+b}$ 4. **Verification of inequality:** - To show $\text{HM} \leq \text{GM}$: $$\frac{2ab}{a+b} \leq \sqrt{ab}$$ Multiply both sides by $a+b$: $$2ab \leq \sqrt{ab}(a+b)$$ Square both sides: $$4a^2b^2 \leq ab(a+b)^2$$ Divide both sides by $ab$ (since $a,b>0$): $$\cancel{ab}4a b \leq \cancel{ab}(a+b)^2 \Rightarrow 4ab \leq (a+b)^2$$ Expand right side: $$4ab \leq a^2 + 2ab + b^2$$ Rearranged: $$0 \leq a^2 - 2ab + b^2 = (a-b)^2$$ Which is always true. - Similarly, to show $\text{GM} \leq \text{AM}$: $$\sqrt{ab} \leq \frac{a+b}{2}$$ Square both sides: $$ab \leq \frac{(a+b)^2}{4}$$ Multiply both sides by 4: $$4ab \leq (a+b)^2$$ Which is the same as above and true. 5. **Conclusion:** The relation $$\frac{2}{\frac{1}{a} + \frac{1}{b}} \leq \sqrt{ab} \leq \frac{a+b}{2}$$ holds for any two unequal positive numbers $a$ and $b$. **Final answer:** $$\boxed{\text{HM} \leq \text{GM} \leq \text{AM}}$$