Review question

# Can we prove these inequalities involving $a, b, c$ and $d$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7865

## Solution

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 $a$, $b$, $c$, $d$: $$$ab\le \left(\frac{a+b}{2}\right)^2, \label{eq:1}$$$ and $$$cd\le \left(\frac{c+d}{2}\right)^2,\label{eq:2}$$$

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.