Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.4 AIME Prove it North Macedonia

Prove that $a^3 + b^3 + 3abc > c^3$ , where $a$ , $b$ , $c$ are the sides of a triangle.

Solution

Let $a$ , $b$ , $c$ be the sides of a triangle. From the triangle inequality $a + b > c$ , we have
$\begin{aligned} a^3 + b^3 + 3abc &= (a+b)(a^2 - ab + b^2) + 3abc \\&> c \cdot (a^2 - ab + b^2) + 3abc \\ &= c \cdot (a^2 - ab + b^2 + 3ab) \\ &= c \cdot (a^2 + 2ab + b^2) \\ &= c \cdot (a+b)^2 \\ &> c \cdot c^2 = c^3 \end{aligned}$

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