Showing that at least one of the equations has two real roots

Question

Let's suppose that $b_1, b_2, c_1, c_2$ are real numbers. We know that $b_1b_2=2(c_1+c_2)$. The task is to prove that at least one of the equations $x^2+b_1x+c_1=0$, $x^2+b_2x+c_2=0$ has two real roots. I tried with $b_1^2-4c_1>0$ or $b_2^2-4c_2>0$ but with no success. However, I have access to a solution (which is probably correct but I don't know why). The solution says: $\Delta_1 + \Delta_2 \ge 0$ and the remaining part is easy (solving this inequality proves that at least one of the equations has two real roots according to the solution), but what I don't understand is the inequality $\Delta_1 + \Delta_2 \ge 0$ - how do we know that the sum is going to be positive or zero? Imagine a situation when $\Delta_1$ =18 and $\Delta_2=-20$. In such case $\Delta_1 + \Delta_2 \ge 0$ is false but still one equation has two real roots (because $\Delta_1$ =18).

Answer 1

if you problem allow this well know?

we say $x^2+bx+c=0$ such $\Delta=b^2-4c=0$,have two real roots.

I think we can use two case

case 1: if $\Delta_{1}=\Delta_{2} =0$,then it is clear

case 2: WLOG

If $$\Delta_{1}< 0,\Delta_{2}\le 0\Longrightarrow b^2_{1}< 4c_{1},b^2_{2}\le 4c_{2}$$ then we have $$b^2_{1}+b^2_{2}<4c_{1}+4c_{2}$$ and $$b^2_{1}+b^2_{2}\ge 2b_{1}b_{2}$$ so we have $$2b_{1}b_{2}< 4(c_{1}+c_{2})$$ impossible