Can we prove the Arithmetic and Geometric Mean Inequality?

Can we prove the Arithmetic and Geometric Mean Inequality?

Question

1. Prove that, if $a$ and $b$ are real, $\frac{1}{2}(a^2+b^2)\geq ab.$

---

1. Deduce that, if $a$,$b$ and $c$ are real, $a^2+b^2+c^2\geq ab+bc+ca.$

---

1. Show that $a+b+c$ is a factor of $a^3+b^3+c^3-3abc$ and find the other factor of this expression.

---

1. Deduce that, if $p$, $q$ and $r$ are positive, $\frac{1}{3}(p+q+r)\geq (pqr)^{\frac{1}{3}}.$

Suggestion

What happens if we bring everything to the left-hand side and try to factorise?

Can we use the first part of the question here?

Could we use earlier parts to help? Would putting $p = a^3, q = b^3, r = c^3$ be more convenient?

Solution

The inequality holds if and only if $a^2+b^2\geq 2ab$, which is true if and only if $a^2- 2ab +b^2\geq 0.$

This is true, since $a^2-2ab+b^2=(a-b)^2,$ and any square is either zero or positive. Hence $\frac{1}{2}(a^2+b^2)\geq ab.$

This is called the arithmetic mean-geometric mean inequality, or the AM-GM inequality, for two variables.

The quantity $\frac{1}{2}(x + y)$ is the arithmetic mean of $x$ and $y$, and (if $x$ and $y$ are positive) $\sqrt{xy}$ is the geometric mean of $x$ and $y$.

So we have established that the arithmetic mean is always greater than or equal to the geometric mean.

This inequality is extremely useful for solving many mathematics problems.

We can use the result in (i) three times for

$a$

and

$b$

,

$b$

and

$c$

and

$c$

and

$a$

, namely

$$\begin{align*} \frac{1}{2}(a^2+b^2) &\geq ab, \\ \frac{1}{2}(b^2+c^2) &\geq bc, \\ \frac{1}{2}(c^2+a^2) &\geq ca. \end{align*}$$

By adding them up we get $\frac{1}{2}(a^2+b^2)+\frac{1}{2}(b^2+c^2)+\frac{1}{2}(c^2+a^2)\geq ab+bc+ca,$ which implies $a^2+b^2+c^2\geq ab+bc+ca.$

We can use the Factor Theorem here.

The Factor Theorem says that the polynomial $f(x)$ has a factor $x-p$ if and only if $f(p)=0$.

Let’s put $f(a) = a^3+b^3+c^3-3abc$, so $f(a)$ is a degree $3$ polynomial in $a$. Thus $a+b+c$ will be a factor of $f(a)$ if and only if $f(-b-c) = 0$.

Now $f(-b-c) = (-b-c)^3+b^3+c^3 -3(-b-c)bc$ $= -b^3-3b^2c-3bc^2-c^3+b^3+c^3+3b^c+3c^2b =0.$

So $a+b+c$ does divide $a^3+b^3+c^3-3abc$.

We need the other factor, which must include $a^2 + b^2 + c^2$, because we need to create $a^3 + b^3 + c^3$.

Let’s say that $(a+b+c)(a^2+b^2+c^2+g)=a^3+b^3+c^3-3abc$, where $g$ is a polynomial in $a, b$ and $c$.

So $a^3+b^3+c^3+ab^2+ac^2+ba^2+ca^2+cb^2 +(a+b+c)g = a^3+b^3+c^3-3abc$ ,

which is true if and only if

$ab^2+ac^2+ba^2+ca^2+cb^2 +3abc +(a+b+c)g =0,$ or $(ab+bc+ca)(a+b+c) + (a+b+c)g=0.$

So $g = -(ab+bc+ca)$, and $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2+-ab-bc-ca).$

Let $a=p^{\frac{1}{3}}$, $b=q^{\frac{1}{3}}$ and $c=r^{\frac{1}{3}}$.

Then in terms of $a$, $b$ and $c$, we have to prove that $a^3+b^3+c^3\geq 3abc.$

By (iii) we know that $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ and this is greater or equal to $0$ since, by (ii), $a^2+b^2+c^2- ab-bc-ca\geq 0$, and $a+b+c\geq 0$, as $a$, $b$ and $c$ are positive.

Therefore, $a^3+b^3+c^3-3abc\geq 0,$ which implies $a^3+b^3+c^3\geq 3abc.$

Hence, $\frac{1}{3}(p+q+r)\geq (pqr)^{\frac{1}{3}}.$

This is the arithmetic mean-geometric mean (AM-GM) inequality for three variables: again, the arithmetic mean is always greater than or equal to the geometric mean.

What happens if we try to generalise to more variables?