Q15 (a) (i)

Criteria	Marks
Provides correct solution	2
Attempts to use appropriate expressions for $x$ and $y$	1

Sample answer:

$\sqrt{abc}=\sqrt{(ab)c}\le\frac{\sqrt{a^2b^2}+c}{2}=\frac{\frac{a^2+b^2}{2}+c}{2}=\frac{a^2+b^2+2c}{4}$

Answers could include:

Let $x=\frac{a^2+b^2}{2}$ and $y=c$.

$\sqrt{\frac{a^2+b^2}{2}\cdot c}\le\frac{\frac{a^2+b^2}{2}+c}{2}$
$\sqrt{\frac{a^2c+b^2c}{2}}\le\frac{a^2+b^2+2c}{4}$

Let $m=a^2c$ and $n=b^2c$.

$\sqrt{mn}=\sqrt{a^2c\cdot b^2c}=abc$
$\sqrt{\frac{m+n}{2}}=\frac{a^2c+b^2c}{2}$
$\therefore abc\le\frac{a^2c+b^2c}{2}$
$\therefore\sqrt{abc}\le\sqrt{\frac{a^2c+b^2c}{2}}\le\frac{a^2+b^2+2c}{4}$