Show that at least one of the equations $x^2+b_1x+c_1 = 0$ and $x^2+b_2x+c_2 = 0$ has two real root. Given that $b_1b_2 = 2(c_1+c_2)$. $b_1$,$b_2$,$c_1$,$c_2$ are real number.
Show that at least one of the equations $$x^2+b_1x+c_1 = 0$$ and $$x^2+b_2x+c_2 = 0$$ has two real root.
 A: Suppose both equations have no real roots. Then $b_1^2\lt 4c_1$ and $b_2^2\lt 4c_2.$ By multiplying these inequalities and combining with the given condition we can obtain $(c_1+c_2)^2\lt 4c_1c_2,$ which implies $$(c_1-c_2)^2\lt 0.$$ Is that possible?
A: If not, $b_i^2<4c_i$ so $c_i>0$ and $4(c_1+c_2)^2=b_1^2b_2^2<16c_1c_2\le 4(c_1+c_2)^2$, contradicting the AM-GM inequality.
A: Hint:   by AM-GM:
$$
\Delta_1+\Delta_2=b_1^2-4c_1 + b_2^2-4c_2 \ge 2 |b_1b_2|-4(c_1+c_2) \ge 2 \big(b_1b_2 - 2(c_1+c_2)\big) = 0
$$
A: For the equations, we have the discriminants $$D_1 = b_1^2 - 4c_1,\ D_2 = b_2^2 - 4c_2 \\ \implies D_1+D_2 = b_1^2 + b_1^2 - 2\cdot 2(c_1 + c_2) = b_1^2 + b_1^2 - 2b_1b_2 = (b_1-b_2)^2 \geq 0$$
Therefore, if no real solutions exist to both equations i.e. $D_1, D_2 < 0$ we arrive at a contradiction. 
A: We want to prove that it is not possible that 
$$b_1^2 - 4c_1 < 0 ~\text{and}~b_2^2 - 4c_2 < 0$$
are both satisfied at the same time.
Suppose that these inequalities are satisfied.
From the assumption $b_1b_2 = 2(c_1+c_2)$, it follows that:
$$4c_1 = 2b_1b_2-4c_2.$$
Then from $b_1^2 - 4c_1 < 0$ we get:
$$b_1^2-2b_1b_2+4c_2 < 0.$$
Moreover, from $b_2^2 - 4c_2 < 0$ we get that:
$$4c_2 > b_2^2.$$
Joining the last two results, one gets:
$$b_1^2-2b_1b_2+b_2^2 < b_1^2-2b_1b_2+4c_2 < 0,$$
or equivalently
$$(b_1-b_2)^2 < b_1^2-2b_1b_2+4c_2 < 0,$$
which implies that
$$(b_1-b_2)^2 < 0,$$
and hence we arrived to a contradiction. Thus, initial supposition must be false.
