Expanding the LHS of the above equation yields: $2x^2 + (a+b+c+d)x + (ad+bc) = 0 \Rightarrow x = \frac{-(a+b+c+d) \pm \sqrt{(a+b+c+d)^2 -8(ad+bc)}}{4}$ (i). The roots of (i) will be real $\Leftrightarrow$ the discriminant $(a+b+c+d)^2 -8(ad+bc) \ge 0$. Let $a = b+p, c = d+q$ for real $p,q \ge 0$ which after substitution into the discriminant gives us: $(2b + 2d + p + q)^2 - 8[(b+p)d + b(d+q)] = 4(b+d)^2 + 4(b+d)(p+q) + (p+q)^2 - 8(2bd+pd+bq) = 4(b-d)^2 + 4(b-d)(p+q) + (p+q)^2 = [2(b-d)+(p+q)]^2 \ge 0$ which is indeed non-negative and therefore yields real roots for $a,b,c,d \in \mathbb{R}$ (where $a \ge b, c \ge d$). QED