Q15 (a) (ii)

$a^2+9b^2\ge 6ab \qquad \text{I}$
$b^2+9c^2\ge 6bc \qquad \text{II}$
$a^2+9c^2\ge 6ac \qquad \text{III}$

By Adding I, II and III

$a^2+9b^2+b^2+9c^2+a^2+9c^2\ge 6ab+6bc+6ac$
$2a^2+10b^2+18c^2\ge 6(ab+bc+ac)$
$a^2+5b^2+9c^2\ge 3(ab+bc+ac)$

1 – setting up the 3 inequalities or equivalent merit
1 – correct setting and result