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The AM–GM inequality

Exercise 11 gave a geometric proof that the arithmetic mean of two positive numbers $a$ and $b$ is greater than or equal to their geometric mean. We can also prove this algebraically, as follows.

Since $a$ and $b$ are positive, we can define $x = \sqrt{a}$ and $y = \sqrt{b}$ . Then

$\begin{align*} (x-y)^2 \geq 0 \quad&\implies\quad x^2+y^2-2xy \geq 0 \\ &\implies\quad \dfrac{x^2+y^2}{2} \geq xy \end{align*}$

and so

$\dfrac{a+b}{2} \geq \sqrt{ab}.$

This is called the AM–GM inequality. Note that we have equality if and only if $a = b$ .

Example

Find the range of the function $f(x) = x^2 + \dfrac{1}{x^2}$ , for $x\neq 0$ .

Solution

Using the AM–GM inequality,

$f(x) = x^2 + \dfrac{1}{x^2} \geq 2\sqrt{x^2 \times \dfrac{1}{x^2}} = 2.$

So the range of $f$ is contained in the interval $[2,\infty)$ . Note that $f(1) = 2$ and that $f(x) \to \infty$ as $x \to \infty$ . Since $f$ is continuous, it follows that, for each $y \geq 2$ , there exists $x \geq 1$ with $f(x) = y$ . Hence, the range of $f$ is the interval $[2,\infty)$ .

Exercise 15

Find the arithmetic, geometric and harmonic means of $3,4,5$ and write them in ascending order.
Prove that the harmonic mean of two positive real numbers $a$ and $b$ is less than or equal to their geometric mean.

The AM–GM inequality can be generalised as follows. If $a_1,a_2,\dots,a_n$ are $n$ positive real numbers, then

$\dfrac{a_1 + a_2 + \dots + a_n}{n} \geq \sqrt[n]{a_1a_2\dots a_n}.$

The next exercise provides a proof of this result.

Exercise 16

Find the maximum value of the function $f(x) = \log_e x - x$ , for $x > 0$ . Hence, deduce that $\log_e x \leq x-1$ , for all $x > 0$ .
Let $a_1,a_2,\dots,a_n$ be positive real numbers and define $A = \dfrac{a_1 + a_2 + \dots + a_n}{n}$ .
By successively substituting $x = \dfrac{a_i}{A}$ , for $i=1,2,\dots,n$ , into the inequality from part (a) and summing, show that $\log_e\Bigl(\dfrac{a_1a_2\dots a_n}{A^n}\Bigr) \leq 0.$
By exponentiating both sides of the inequality from part (b), derive the generalised AM–GM inequality.

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