sqing wrote: Wellingson wrote: 2) Let $a, b, c$ be real numbers such that $$a+\frac{b}{c}=b+\frac{c}{a}=c+\frac{a}{b}=1$$a) Prove that $ab+bc+ca=0$ and $a+b+c=3$. b) Prove that $|a|+|b|+|c|< 5$ What is ONEM ? Thanks. Olimpiada Nacional Escolar de Matemática (ONEM) [Peru]

The existence of the fractions: $a\ne0;b\ne0;c\ne0$. $a+\dfrac{b}{c}=1\Longrightarrow ac=c-b\quad(1)$; $b+\dfrac{c}{a}=1\Longrightarrow ab=a-c\quad(2)$; $c+\dfrac{a}{b}=1\Longrightarrow bc=b-a\quad(3)$. Adding the relations $(1),(2),(3)$ results: $ab+ac+bc=0\quad(4)$. Adding the initial equations results: $a+b+c+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{a}{b}=3\quad(5)$. $\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{a}{b}=\dfrac{ab^2+bc^2+ca^2}{abc}$. Using the relations$(4),(1),(3)$ we obtain: $ab^2+bc^2+ca^2=ab^2+bc^2+ca^2-b(ab+ac+bc)=c(bc+a^2-ab-b^2)=$ $=c[b(c-b)+a(a-b)]=c(b\cdot ac-a\cdot bc)=0$. Replacing in $(5)$ results: $a+b+c=3$. $a,b,c$ cannot be all positive. If we assume $a,b,c>0$ results $a=1-\dfrac{b}{c}<1$ and similarly $b<1,c<1$, hence $a+b+c<3$, contradiction. Assume $a<0$. $a+\dfrac{b}{c}=1\Longrightarrow \dfrac{b}{c}=1-a$; $b+\dfrac{c}{a}=1\Longrightarrow \dfrac{c}{a}=1-b$; $c+\dfrac{a}{b}=1\Longrightarrow \dfrac{a}{b}=1-c$. Multiplying the relations results: $(1-a)(1-b)(1-c)=1\Longrightarrow abc=ab+ac+bc-(a+b+c)=-3\Longrightarrow$ $\Longrightarrow bc>0$, hence $b,c$ have the same sign. From $b<0,c<0$ results $a+b+c<0$, contradiction. Results: $b>0,c>0$. Denote $a=-A, b+c=S>0,bc=P>0$. $|a|+|b|+|c|=A+b+c$. $a+b+c=3\Longrightarrow A=S-3>0\Longrightarrow S>3$. $ab+ac+bc=0\Longrightarrow A=\dfrac{P}{S}$. $A=S-3=\dfrac{P}{S}\Longrightarrow P=S^2-3S$. Using AM-GM, results: $P\le\dfrac{S^2}{4}\Longrightarrow S^2-3S\le\dfrac{S^2}{4}\Longrightarrow S\le4$. $A+b+c=S-3+S=2S-3\le5$. We will prove: the equality cannot occur. For $S=4$ results $P=4; b=c=2; a=-A=-1$. But for these values $a+\dfrac{b}{c}=0$, in contradiction with the initial condition $a+\dfrac{b}{c}=1$. Results: $|a|+|b|+|c|=A+b+c<5$.

Peru National School Mathematics Olympiad

$$ab=a-c,bc=b-a,ca=c-b, ab+ac+bc=0$$$$ab^2+bc^2+ca^2=ab^2+bc^2+ca^2-b(ab+ac+bc)$$$$=c(bc+a^2-b^2-ab)=c(b(c-b)+a(a-b))=0$$$$3=a+b+c+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{a}{b}=a+b+c+\dfrac{ab^2+bc^2+ca^2}{abc}=a+b+c$$$$a+b+c=3$$

Let $a, b, c$ be real numbers such that $a+\frac{b}{c}=b+\frac{c}{a}=c+\frac{a}{b}=1.$ Prove that $$abc = -3.$$and $$|a|+|b|+|c|\leq 1+\frac{8\sqrt{2}}{3}.$$

Interesting fact: The system has three solutions. They are the cyclic permutations of $\left(1+2\cos\frac{2\pi}{9},1+2\cos\frac{4\pi}{9},1+2\cos\frac{8\pi}{9}\right)$.

Let $a, b, c$ be real numbers such that $a+\frac{b}{c}=b+\frac{c}{a}=c+\frac{a}{b}=1.$ Prove that： $a,b,c$ are roots of $x^3-3x^2+3=0.$

Math5000 wrote: 2) Let $a, b, c$ be real numbers such that $$a+\frac{b}{c}=b+\frac{c}{a}=c+\frac{a}{b}=1$$a) Prove that $ab+bc+ca=0$ and $a+b+c=3$. b) Prove that $|a|+|b|+|c|< 5$