Prove that if $abc\neq 0$ then at least one of the equations $ax^2+2bx+c$, $bx^2+2cx+a$, $cx^2+2ax+b$ has root.
Source: All-Russian Math Olympiad 1994.
My sketch of proof: The condition $abc\neq 0$ is equivalent to $a,b,c\neq 0$.
Suppose that none of these equation have root. Then we have the following three inequalities: $b^2-ac<0, c^2-ab<0, a^2-bc<0.$ Adding all these three inequalities we get: $a^2+b^2+c^2<ab+ac+bc$. But we can easily show that $a^2+b^2+c^2\geq ab+bc+ac$ which leads to contradiction.
However, this approach seems to me quite weird because I have not used the condition that $abc\neq 0.$
After some time I got the correct solution which is the following: Since $b^2<ac, c^2<ab, a^2<bc$ we get that $(abc)^2<(abc)^2$ and with $abc\neq 0$ this leads to contradiction.
But what's wrong with my approach?
Yes you have used $abc\ne0$. Out of that condition, say $a=0$ you don't have quadratic equation $$ax^2+bx+c=0$$ so you can't say $b^2-ac<0$