$P(x)=x^2+ax+b$ and $Q(x)=x^2+a_1x+b_1$ and $P(x)\cdot Q(x)$ has at most one real root then prove that the claims given below are true. 
If $P(x)=x^2+ax+b$ and $Q(x)=x^2+a_1x+b_1$ where $a,b,a_1,b_1\in\mathbb{R}$ and equation $P(x)\cdot Q(x)=0$ has at most one real root then prove that the claims given below are true.

$1.$ $(1+a+b)(1+a_1+b_1)>0$
$2.$ $(1+a+b)>0$
$3.$ $\displaystyle\frac{1+a+b}{1+a_1+b_1}>0$
My work:
Let $P(x)\cdot Q(x)=f(x)$ then $f(x)$ has $0$ real roots as it can't have only $1$ real root.
$$f(x)=x^4+x^3(a+a_1)+x^2(aa_1+b+b_1)+x(ab_1+ba_1)+bb_1$$ Now, since the leading coefficient of $f(x)$ is $+ve$ it means that $f(x)>0\:\:\forall x\in\mathbb{R}$ $\implies f(1)>0$
$$f(1)=1+(a+a_1)+(aa_1+b+b_1)+(ab_1+ba_1)+bb_1$$ $$=(1+a+b)(1+a_1+b_1)>0$$
Since $f(x)$ has $0$ real roots, it means that both $P(x)$ and $Q(x)$ have $0$ real roots $\implies$ both $P(1)>0$ and $Q(1)>0$ as both of their leading coefficients are positive.
Or individually,
$$
\begin{align}
(1+a+b)>0\\
(1+a_1+b_1)>0\\
\end{align}$$ It $\implies$ $$\frac{1+a+b}{1+a_1+b_1}>0$$
Is my solution correct$?$

When it says both have at most one real root, it means $a^2-4b\le0$ and $a_1^2-4b_1\le0$.
Also, both inequalities cannot be equality, because they will have two real roots, or if both inequalities are equality, then $a=a_1$ and $b=b_1$.
Then $a+b+1\ge a+a^2/4+1=(a/2+1)^2\ge 0$
BUT
Because it can be $a=a_1=-2$ and $b=b_1=1$ with the only root of $1$, then it is a counterexample for all claims (1,2,3) where they are respectively 0, 0, and ambiguous.
