Well done Freddie Manners age 11, Packwood Haugh School, Shropshire and Danny Ng, age 16, Milliken High School, Canada,

Is the mean of the squares of two numbers greater than, or less than, the square of their means? Let the two numbers be $p$ and $q$

Square of mean $= \frac{(p + q)^{2}}{4} = \frac{p^{2} + 2 pq + q^{2}}{4}$

Mean of squares = $\frac{p^{2} + q^{2}}{2}$

$\frac{p^{2} + q^{2}}{2} - \frac{p^{2} + 2 pq + q^{2}}{4} = \frac{p^{2} - 2 pq + q^{2}}{4} = \frac{(p - q)^{2}}{4} ≧ 0 .$

Note that this difference, $\frac{(p - q)^{2}}{4}$ , is zero if $p = q$ and positive for all other choices of $p$ and $q$ . So if the numbers are equal then the mean of the squares is equal to the square of the mean. Otherwise the mean of the squares is greater than the square of the mean.