Library / Macedonian Mathematical Competitions /12 of 73

Macedonian Mathematical Competitions

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
$\begin{aligned} a^3 + b^3 + 3abc &= (a+b)(a^2 - ab + b^2) + 3abc \\&> c \cdot (a^2 - ab + b^2) + 3abc \\ &= c \cdot (a^2 - ab + b^2 + 3ab) \\ &= c \cdot (a^2 + 2ab + b^2) \\ &= c \cdot (a+b)^2 \\ &> c \cdot c^2 = c^3 \end{aligned}$

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.