Prove that, if \(a\), \(b\) are real, \[ab\le \left(\frac{a+b}{2}\right)^2, ... \label{eq:star}\tag{$*$} \]

Since \(a\) and \(b\) are real, we have that \[(a-b)^2\ge 0.\] Expanding and rearranging, we have that \[\begin{align*} &&(a-b)^2&\ge 0,&&\quad \\ \iff\quad&&a^2+b^2-2ab&\ge 0, \\ \iff\quad&&(a+b)^2-4ab&\ge 0, \\ \iff\quad&&4ab&\le (a+b)^2,\\ \end{align*}\] therefore, \[\begin{align*} ab\le \left(\frac{a+b}{2}\right)^2. \end{align*}\]

Note also that we have equality on the first line of this argument if and only if \(a=b\), so the final inequality \(\eqref{eq:star}\) is also true if and only if \(a=b\).

Alternatively, we could have begun with the given statement \(\eqref{eq:star}\) and worked our way backwards to \((a-b)^2\ge0\), ensuring that we use “if and only if” (\(\iff\)) arguments the whole way.

… and deduce that, if \(a\), \(b\), \(c\), \(d\) are positive, \[\begin{equation*} abcd\le \left(\frac{a+b+c+d}{4}\right)^4,\label{eq:dagger}\tag{$\dagger$} \end{equation*}\]

with equality only when all numbers are equal.

We may apply inequality \(\eqref{eq:star}\) to any two pairs of \(a\), \(b\), \(c\), \(d\): \[\begin{equation} ab\le \left(\frac{a+b}{2}\right)^2, \label{eq:1} \end{equation}\] and \[\begin{equation} cd\le \left(\frac{c+d}{2}\right)^2,\label{eq:2} \end{equation}\]

with equality in \(\eqref{eq:1}\) if and only if \(a=b\), and equality in \(\eqref{eq:2}\) if and only if \(c=d\).

Since both sides of \(\eqref{eq:1}\) and \(\eqref{eq:2}\) are positive, we have that \[abcd\le \left(\frac{a+b}{2}\right)^2 \left(\frac{c+d}{2}\right)^2 = \left[ \left(\frac{a+b}{2}\right) \left(\frac{c+d}{2}\right) \right]^2,\] with equality if and only if \(a=b\) and \(c=d\), and we may further apply \(\eqref{eq:star}\) to the right hand side of this to find that \[abcd\le \left(\frac{\dfrac{a+b}{2}+\dfrac{c+d}{2}}{2}\right)^4= \left(\frac{a+b+c+d}{4}\right)^4\] with equality if and only if, in addition to \(a=b\) and \(c=d\), we further have \((a+b)/2=(c+d)/2\), that is \(a=c\). So we have equality here if and only if \(a=b=c=d\), as required.

Note that “only when” is the same as saying “only if”. We have gone further and shown “when and only when”, or “if and only if”.

By giving \(d\) a suitable value in terms of \(a\), \(b\), \(c\), or otherwise, prove that, if \(a\), \(b\), \(c\) are positive, \[abc\le \left(\frac{a+b+c}{3}\right)^3.\]

Approach 1

We want to modify our equation in such a way as to make it look more like \(\eqref{eq:dagger}\). It would be helpful if we could choose a value of \(d\) that would turn the expression in the bracket into the one we want. Hence we require \(d\) such that

\[\frac{a+b+c}{3}=\frac{a+b+c+d}{4},\] so multiplying by \(4\) and subtracting \(a+b+c\) from both sides, we find that we must choose \(d=\dfrac{a+b+c}{3}\). Substituting this into \(\eqref{eq:dagger}\), we get \[abc\left(\frac{a+b+c}{3}\right)\le \left(\frac{a+b+c+\dfrac{a+b+c}{3}}{4}\right)^4 = \left(\frac{a+b+c}{3}\right)^4\]

As \(a\), \(b\) and \(c\) are positive, we can divide both sides by \(\dfrac{a+b+c}{3}\) to get the expression we required:

\[abc\le \left(\frac{a+b+c}{3}\right)^3.\]

Approach 2

As both sides of the equation are positive, another way to turn the cube back into a fourth power is to raise the inequality to the power of \(\frac{4}{3}\): \[\begin{align*} &&abc&\le \left(\dfrac{a+b+c}{3}\right)^3&&\quad\\ \iff\quad&&(abc)^{4/3}&\le \left(\dfrac{a+b+c}{3}\right)^4. \end{align*}\]

If we compare this inequality to the previous result, we see that it may be helpful to choose \(d\) to make the left hand sides match up.

We want \((abc)^{4/3}=abcd\), so we require \(d=(abc)^{1/3}\). If we substitute this into \(\eqref{eq:dagger}\), we deduce that \[\begin{align*} && abc(abc)^{1/3} &\le\left(\frac{a+b+c+(abc)^{1/3}}{4}\right)^4&&\quad \\ \Longrightarrow \quad&& (abc)^{1/3} &\le\frac{a+b+c+(abc)^{1/3}}{4} \\ \Longrightarrow \quad&& (abc)^{1/3} - \frac{(abc)^{1/3}}{4}&\le\frac{a+b+c}{4} \\ \Longrightarrow \quad&& (abc)^{1/3} &\le\frac{a+b+c}{3} \\ \Longrightarrow \quad&& abc &\le\left(\frac{a+b+c}{3} \right)^3. \end{align*}\]

This question has covered the first few cases of the arithmetic mean–geometric mean (AM–GM) inequality. This states that if \(a_1\), …, \(a_n\) are positive real numbers, then \[\frac{a_1+\cdots+a_n}{n}\ge \sqrt[n]{a_1\dots a_n},\] that is, their arithmetic mean is at least as large as their geometric mean.

One of the proofs of this inequality is exactly as shown in this question: one proves the result first for \(n=2^k\) for \(k=1\), \(2\), …, using the method of the first part of the question, and then one fills in the gaps for \(2^{k-1}<n<2^k\) using the technique of the second part.