2023TrialShort answer

Q14 (b) (iii)

3 marks

Question

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

(1+a^3)(1+b^3)(1+c^3)\ge\left(\frac{ab+bc+ca+1}{2}\right)^3.

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Solution

\frac{a^2}{1+a^3}+\frac{b^2}{1+b^3}+\frac{1}{1+c^3} \ge \frac{3abc}{\sqrt[3]{(1+a^3)(1+b^3)(1+c^3)}} \tag{1}

\frac{a^3}{1+a^3}+\frac{1}{1+b^3}+\frac{c^3}{1+c^3} \ge \frac{3abc}{\sqrt[3]{(1+a^3)(1+b^3)(1+c^3)}} \tag{2}

\frac{1}{1+a^3}+\frac{b^3}{1+b^3}+\frac{c^3}{1+c^3} \ge \frac{3abc}{\sqrt[3]{(1+a^3)(1+b^3)(1+c^3)}} \tag{3}

\frac{1}{1+a^3}+\frac{1}{1+b^3}+\frac{1}{1+c^3} \geq \frac{3}{\sqrt[3]{(1+a^3)(1+b^3)(1+c^3)}}

\textcircled{1}+\textcircled{2}+\textcircled{3}+\textcircled{4}

6\geq \frac{3\sqrt[3]{(ab+bc+ca+1)}}{\sqrt[3]{(1+a^3)(1+b^3)(1+c^3)}}

\sqrt[3]{(1+a^3)(1+b^3)(1+c^3)} \geq \left(\frac{ab+bc+ca+1}{2}\right)

(1+a^3)(1+b^3)(1+c^3) \geq \left(\frac{ab+bc+ca+1}{2}\right)^3

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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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