UkraineMO

Algebra Difficulty 4.6 AIME Prove it Ukraine

Real nonzero numbers $a$ , $b$ , $c$ , $d$ satisfy the conditions $a^3 + b^3 + c^3 + d^3 = 0$ and $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \neq 0$ . Prove that $a + b + c + d \neq 0$ .

Solution

Solution. We will use method from contradiction. Suppose $a + b + c + d = 0$ . Thus $a + b = -(c + d)$ . Therefore, we can obtain such equalities:
$(a + b)^3 = -(c + d)^3 \Leftrightarrow a^3 + b^3 + 3ab(a + b) = -c^3 - d^3 - 3cd(c + d) \Leftrightarrow \\ 3ab(a + b) + 3cd(c + d) = 0 \Leftrightarrow -3ab(c + d) - 3cd(a + b) = 0 \Leftrightarrow \\ ab(c + d) + cd(a + b) = 0 \Leftrightarrow abc + abd + acd + bcd = 0.$
That contradicts the conditions since:
$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \neq 0 \Leftrightarrow \frac{abc + abd + acd + bcd}{abcd} \neq 0 \Leftrightarrow abc + abd + acd + bcd \neq 0.$

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.