Because $a_1$, $b_1$ and $c_1$ are all positive, therefore $-\frac{b_1}{a_1} < 0$ and $\frac{c_1}{a_1}>0$. By Vieta's theorem, we find the sum of these two roots is negative, but their product is positive. This means both roots are negative. By the same reasoning the roots of the $2^{nd}$ equation must be both negative too. Because $a_1a_2$, $b_1b_2$ and $c_1c_2$ are all positive, we only need to show that the equation $a_1a_2x^2+b_1b_2x+c_1c_2=0$ is solvable in real numbers. If so, by the same reasoning, its roots must be both negative. $$\Delta = (b_1b_2)^2 - 4(a_1a_2)(c_1c_2) = b_1^2b_2^2-a_1c_1(4a_2c_2)$$ Because $a_2x^2+b_2x+c_2=0$ is solvable in real numbers, it must be true that $b_2^2\ge 4a_2c_2$. It follows that $$ b_1^2b_2^2-a_1c_1(4a_2c_2) \ge b_1^2(4a_2c_2)-a_1c_1(4a_2c_2)=4a_2c_2(b_1^2-a_1c_1)>0$$ The last inequality holds because $a_1x^2+b_1x+c_1=0$ has two real roots which means $b_1^2\ge 4a_1c_2$.