2021HSCShort answer

Q15 (a) (ii)

2 marks

Question

Using part (i), or otherwise, show that for all non-negative real numbers $a,b,c$ ,
$\sqrt{abc}\le\frac{a^2+b^2+c^2+a+b+c}{6}.$

Try the question before revealing the answer.

Solution

Criteria	Marks
Provides correct solution	2
Permutes the result in (i), or equivalent merit	1

Sample answer:

$\sqrt{abc}\le\frac{a^2+b^2+2c}{4},\quad\sqrt{acb}\le\frac{a^2+c^2+2b}{4},\quad\sqrt{bca}\le\frac{b^2+c^2+2a}{4}$

Adding:

$3\sqrt{abc}\le\frac{2a^2+2b^2+2c^2+2a+2b+2c}{4}$
$3\sqrt{abc}\le\frac{a^2+b^2+c^2+a+b+c}{2}$
$\therefore\sqrt{abc}\le\frac{a^2+b^2+c^2+a+b+c}{6}\quad\text{as required}$

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Proof

The Nature of ProofUse formal proof language, contradiction, counter-examples and rigorous arguments involving numbers and inequalities.

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