Is the AM-GM Inequality True for Four Numbers?

For four nonnegative numbers, the arithmetic mean is at least the geometric mean. In symbols, for $a,b,c,d\geq 0$,

$$\frac{a+b+c+d}{4}\geq \sqrt[4]{abcd}.$$

The inequality is strict unless all four numbers are equal. A convenient proof groups the numbers into two pairs and uses the two-number AM-GM inequality twice.

Apply AM-GM to two pairs

For any nonnegative $x$ and $y$, the two-number AM-GM inequality says

$$\frac{x+y}{2}\geq \sqrt{xy}.$$

Apply it first to $a,b$ and then to $c,d$:

$$\frac{a+b}{2}\geq \sqrt{ab}, \qquad \frac{c+d}{2}\geq \sqrt{cd}.$$

The arithmetic mean of all four numbers is the average of these two pairwise arithmetic means. Therefore,

$$\frac{a+b+c+d}{4} =\frac{1}{2}\left(\frac{a+b}{2}+\frac{c+d}{2}\right) \geq \frac{\sqrt{ab}+\sqrt{cd}}{2}.$$

Now use AM-GM once more, this time on $\sqrt{ab}$ and $\sqrt{cd}$:

$$\frac{\sqrt{ab}+\sqrt{cd}}{2} \geq \sqrt{\sqrt{ab}\,\sqrt{cd}} =\sqrt[4]{abcd}.$$

Combining the two inequalities gives

$$\boxed{\frac{a+b+c+d}{4}\geq \sqrt[4]{abcd}}.$$

This proof works even if one or more numbers are zero. If you are given four positive numbers, the same result applies without needing to consider zero-valued cases.

When does equality hold?

Equality in the first two applications requires $a=b$ and $c=d$. Equality in the last application requires $\sqrt{ab}=\sqrt{cd}$. Together, these conditions mean all four numbers are equal.

Conversely, if $a=b=c=d=k$ for some $k\geq 0$, then both means equal $k$. Thus the exact statement is: the arithmetic mean is greater than or equal to the geometric mean, with equality exactly when all four numbers are equal. If they are not all equal, the inequality is strict.

Worked example

Take the four numbers $1,2,8,16$. Their arithmetic mean is

$$\frac{1+2+8+16}{4}=\frac{27}{4}=6.75.$$

Their product is $1\cdot2\cdot8\cdot16=256$, so their geometric mean is

$$\sqrt[4]{256}=4.$$

Therefore $6.75>4$, as expected: the four numbers are not all equal. The proof also shows how the comparison can be made pair by pair. Pairing $1$ with $2$ and $8$ with $16$ gives

$$\frac{1+2+8+16}{4} \geq \frac{\sqrt{2}+\sqrt{128}}{2} =\frac{9\sqrt{2}}{2} \geq 4.$$

The intermediate bound need not equal the geometric mean; it only needs to be at least as large as it.

Common mistakes

• Writing “greater than” without an equality case: If all four numbers are equal, the two means are equal. Use $\geq$ for the general theorem.
• Forgetting the nonnegative assumption: The usual real-valued geometric mean here is defined for nonnegative inputs. Negative numbers require separate care, since the fourth root of their product may not be real.
• Taking the wrong root: With four numbers, the geometric mean is the fourth root of the product, not the square root.
• Assuming every pairing gives the same intermediate bound: The final AM-GM inequality is independent of how you group the numbers, but the intermediate estimates in this proof can vary with the pairing.

Generalization

For $n$ nonnegative numbers, the AM-GM inequality states that their arithmetic mean is at least the $n$th root of their product:

$$\frac{x_1+x_2+\cdots+x_n}{n} \geq (x_1x_2\cdots x_n)^{1/n}.$$

The four-number result is the case $n=4$. The pairwise proof above is a particularly short way to establish that case from the familiar two-number inequality.