If A, G and H are A.M., G.M. and H.M. respectively between any two unequal positive numbers then prove that (i) A G H and (ii) G^2 = A H

Let $a$ and $b$ be two unequal positive numbers such that $a\ne b$ and $a,b>0$. Let $A$, $G$, and $H$ be the arithmetic mean (A.M.), geometric mean (G.M.), and harmonic mean (H.M.) of $a$ and $b$, respectively.

Definitions

1. Arithmetic Mean (A): $$A=\frac{a+b}{2}$$

2. Geometric Mean (G): $$G=\sqrt{ab}$$

3. Harmonic Mean (H): $$H=\frac{2ab}{a+b}$$

---

(i) Prove that $A>G>H$

Step 1: Prove $A>G$

We need to show that: $$\frac{a+b}{2}>\sqrt{ab}$$

Square both sides (since $a,b>0$, squaring preserves the inequality): $${\left(\frac{a+b}{2}\right)}^2>ab$$

Simplify: $$\frac{(a+b{)}^2}{4}>ab$$ $$(a+b{)}^2>4ab$$ $$a^2+2ab+b^2>4ab$$ $$a^2-2ab+b^2>0$$ $$(a-b{)}^2>0$$

Since $a\ne b$, $(a-b{)}^2>0$ is always true. Thus, $A>G$.

Step 2: Prove $G>H$

We need to show that: $$\sqrt{ab}>\frac{2ab}{a+b}$$

Since $a,b>0$, we can multiply both sides by $\frac{a+b}{\sqrt{ab}}$ (which is positive): $$(a+b)>\frac{2ab}{\sqrt{ab}}$$ $$(a+b)>2\sqrt{ab}$$ $$\frac{a+b}{2}>\sqrt{ab}$$

But this is the same as $A>G$, which we have already proven. Alternatively, we can directly compare $G$ and $H$:

$$G>H⟹\sqrt{ab}>\frac{2ab}{a+b}$$

Divide both sides by $\sqrt{ab}$ (positive): $$1>\frac{2\sqrt{ab}}{a+b}$$

Multiply both sides by $(a+b)$: $$a+b>2\sqrt{ab}$$

This is equivalent to: $$\frac{a+b}{2}>\sqrt{ab}$$

Which again reduces to $A>G$. Thus, $G>H$.

Conclusion for (i)

Combining both results, we have: $$A>G>H$$

---

(ii) Prove that $G^2=A\times H$

Step 1: Express $G^2$, $A$, and $H$ explicitly

$$G^2=(\sqrt{ab}{)}^2=ab$$ $$A=\frac{a+b}{2}$$ $$H=\frac{2ab}{a+b}$$

Step 2: Compute $A\times H$

$$A\times H=\left(\frac{a+b}{2}\right)\times \left(\frac{2ab}{a+b}\right)$$

Simplify: $$A\times H=\frac{(a+b)\times 2ab}{2(a+b)}$$ $$A\times H=ab$$

But $G^2=ab$, so: $$G^2=A\times H$$

---

Final Answer

(i) For any two unequal positive numbers, the arithmetic mean $A$, geometric mean $G$, and harmonic mean $H$ satisfy: $$A>G>H$$

(ii) The relationship between $G$, $A$, and $H$ is: $$G^2=A\times H$$