Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.6 AIME Prove it Ireland

Suppose $a$ , $b$ , $c$ are real numbers such that $a + b + c = 1$ and $a^2 + b^2 + c^2 = 1$ . Prove that $a^3 + b^3 + c^3 \ge \frac{5}{9}$ .

Solution

In the solution to Problem 7 we have seen that $a + b + c = 1$ and $a^2 + b^2 + c^2 = 1$ imply $a^3 + b^3 + c^3 = 1 + 3abc$ . Hence, $a^3 + b^3 + c^3 \ge \frac{5}{9}$ iff $abc \ge -\frac{4}{27}$ . Several proofs of this inequality can be found in the solution to Problem 11.

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