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

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.

- Deduce that, if \(a\),\(b\) and \(c\) are real, \[a^2+b^2+c^2\geq ab+bc+ca.\]

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 \[a^2+b^2+c^2\geq ab+bc+ca.\]

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

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 \[a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2+-ab-bc-ca).\]

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

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?