Prove atleast one of the equations $x^2+b_1x+c_1=0$ and $x^2+b_2x+c_2=0$ has real roots if $b_1b_2=2(c_1+c_2)$

Question

If $b_1b_2=2(c_1+c_2)$, then prove that atleast one of the equations $x^2+b_1x+c_1=0$ and $x^2+b_2x+c_2=0$ has real roots.

$$ \Delta_1.\Delta_2=(b_1^2-4c_1)(b_2^2-4c_2)=b_1^2b_2^2-4b_2^2c_1-4b_1^2c_2+16c_1c_2\\ =4(c_1+c_2)^2-4b_2^2c_1-4b_1^2c_2+16c_1c_2\\ =4c^2_1+4c_2^2+8c_1c_2-4b_2^2c_1-4b_1^2c_2+16c_1c_2\\ =4\big[c^2_1+c_2^2+2c_1c_2-b_2^2c_1-b_1^2c_2+4c_1c_2\big]\\ =4\big[c^2_1+c_2^2+6c_1c_2-b_2^2c_1-b_1^2c_2\big] $$ Why am I not reaching any conclusion from the product of the determinants ?

Answer 1

Say non has real solution, then $$b_1^2<4c_1\;\;\;\wedge \;\;\;b_2^2<4c_2$$ Then $$b_1^2b_2^2<16c_1c_2$$ so $$4(c_1^2+2c_1c_2+c_2^2)<16c_1c_2$$ so$$c_1^2-2c_1c_2+c_2^2<0$$

or $$(c_1-c_2)^2<0$$ A contradiction.

Answer 2

$$ \Delta_1+\Delta_2=(b_1^2-4c_1)+(b_2^2-4c_2)=(b_1-b_2)^2\ge0$$

If both $\Delta_1,\Delta_2<0,$

$$\Delta_1+\Delta_2<0\not\ge0$$

Answer 3

We can assume both $c_1,c_2$ greater than zero.

$b_1b_2=2(c_1+c_2) = 4\dfrac{(c_1+c_2)}{2}$

$b^2_1b^2_2=16\dfrac{(c_1+c_2)^2}{4}\geq 16c_1c_2$

$b^2_1b^2_2 \geq 4c_14c_2$

So if $b^2_1 \lt 4c_1$ , then $b^2_2 \gt 4c_2$ and vice versa.