2023TrialShort answer

Q14 (b) (i)

2 marks

Question

Given $a$ , $b$ , $c$ are positive real numbers, prove that

a^2+b^2+c^2\ge ab+bc+ca.

Try the question before revealing the answer.

Solution

(a-b)^2\ge0\quad\therefore\quad a^2+b^2\ge2ab

Similarly,

a^2+c^2\ge2ac

b^2+c^2\ge2bc

2(a^2+b^2+c^2)\ge2(ab+bc+ca)

\therefore\quad a^2+b^2+c^2\ge ab+bc+ca

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Syllabus alignment

Proof

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

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