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Question:

Show that the harmonic mean is less than or equal to the arithmetic mean, and also less than or equal to the geometric mean, with equality if and only if $a=b$ ; ie., show that $$\frac{a+b}{2} \geq \sqrt{ab} \geq \frac{2ab}{a+b}$$

My attempt,

I've shown the first equality which is $$\frac{a+b}{2}-\frac{2ab}{a+b}=(a-b)^2\geq0$$

My first question, should I write $\frac{(a-b)^2}{2(a+b)}\geq0$ or just leave as $(a-b)^2\geq0$?

For the second,

$$\sqrt{ab}-\frac{2ab}{a+b}=\frac{\sqrt{ab}(a+b)-2ab}{a+b}$$

$$=\frac{a^{\frac{3}{2}} \sqrt{b}+\sqrt{a} b^{\frac{3}{2}}-2ab}{a+b}$$

$$=\frac{\sqrt{ab}(a+b-2\sqrt{ab})}{a+b}$$

I stuck at here. Can anyone explain to me how to solve this question Thanks in advance.

**I found other similar questions being asked in this site. But my question includes harmonic which is not a duplicate.

Mathxx
  • 7,570

2 Answers2

2

From AM-GM, $$\frac {a+b}2\ge\sqrt{ab}\tag{1}$$ Cross-multiplying,

$$\frac 1{\sqrt{ab}}\ge \frac 2{a+b}\\ \frac{\sqrt{ab}}{ab}\ge \frac 2{a+b}$$ $$\sqrt{ab}\ge \frac {2ab}{a+b}\tag{2}$$ Combining $(1),(2)$ gives

$$\color{red}{\frac {a+b}2\ge \sqrt{ab}\ge \frac {2ab}{a+b}}$$

Hypergeometricx
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1

You are almost there with the second part: $$ \frac{\sqrt{ab}(a+b-2\sqrt{ab})}{a+b} \ge 0 $$ because $$ a+b\ge2\sqrt{ab} \quad\Leftrightarrow\quad {a+b\over2}\ge\sqrt{ab} $$ (this has been already proved in the first part). Just like with the AM-GM case, we have equality only when $a=b$.

Alex
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