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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 \(a = b\).


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


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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