### 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

1. Find the arithmetic, geometric and harmonic means of $$3,4,5$$ and write them in ascending order.
2. 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

1. 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 $$x > 0$$.
2. 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 $\log_e\Bigl(\dfrac{a_1a_2\dots a_n}{A^n}\Bigr) \leq 0.$
3. By exponentiating both sides of the inequality from part (b), derive the generalised AM–GM inequality.

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