Pythagorean means

This problem explores the classic ordering of the three Pythagorean means—arithmetic ((A)), geometric ((G)), and harmonic ((H))—for two positive numbers. It’s a beautiful exercise linking inequalities, sequences, and algebraic manipulation, directly relevant to the HSC topic of sequences and series. By working through the proof and the arithmetic progression conditions, you’ll deepen your understanding of mean relationships and how to set up equations for AP/GP.

Problem Statement

For $0<a<b$, define

$$A=\frac{a+b}{2},\quad G=\sqrt{ab},\quad H=\frac{2ab}{a+b}.$$

• Prove $H<G<A$.
• Find $\frac ba$ such that $a,G,A$ is an AP.
• Show $H,G,A$ cannot be an AP unless $a=b$.

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Hints

Use $(\sqrt a-\sqrt b)^2>0$ and $G^2=AH$.

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Solutions

We start by proving the inequality. Using the hint, the square of a difference is positive, which gives us $A>G$. Notice the relationship between $G$, $A$, and $H$: the product $AH$ equals $ab$, which is $G^2$. This tells us that $H$, $G$, $A$ are in geometric progression. Since $G$ is the geometric mean and we already know $A>G$, it follows that $H<G<A$.
To make the argument concrete, we expand the square:

From $(\sqrt a-\sqrt b)^2>0$, get $A>G$. Also

$$AH=\frac{a+b}{2}\cdot\frac{2ab}{a+b}=ab=G^2,$$

so $H,G,A$ form a GP and therefore $H<G<A$.

Next, we set up the condition for $a$, $G$, $A$ to be an arithmetic progression. The definition of an AP gives us the middle term as the average of the first and last, so $2G = a + A$. Substituting the expressions and simplifying leads to an equation in the ratio $x = b/a$.

For $a,G,A$ in AP:

$$2G=a+A \implies 4\sqrt{ab}=3a+b.$$

Let $x=b/a$; then $(x-9)(x-1)=0$, and $x>1$ gives $x=9$.

Finally, suppose $H$, $G$, $A$ are in arithmetic progression. Using the GP relationship $G^2 = AH$, we write $H = G^2/A$. Substituting into $2G = H + A$ gives an equation that forces $A$ and $G$ to be equal, which contradicts the strict inequality unless $a = b$. Thus, they cannot be an AP for distinct positive numbers.

If $H,G,A$ were AP, then $2G=H+A=G^2/A+A$, forcing $A=G$, contradiction for $a\ne b$.

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Takeaways

• The inequality between arithmetic, geometric, and harmonic means can be proven using simple algebraic manipulations and the fact that $G^2 = AH$.
• When dealing with means of two numbers, setting up AP or GP conditions often reduces to solving a quadratic equation in the ratio $b/a$.
• The relationship $H,G,A$ as a GP is a key insight: it ties the means together and makes ordering and AP checks straightforward.

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Further Readings

HSC Distributions, HSC Induction, HSC Integrals, HSC Sequences