Let $f(x)=p(x-a)(x-c)+q(x-b)(x-d)=(p+q)x^2-(p(a+c)+q(b+d))x+pac+qbd$ where $p,q\in \mathbb{R}^+$. Since $a<b\leq c<d$, it is easily seen that $f(a)>0, f(b)\leq 0, f(c)\leq 0, f(d)>0$ where the equalities are only possible when $b=c$. Hence $f(x)$ has a root and so $D\geq 0$. Now the inequality in the problem is equivalent to $D\geq 0$ when $p=\frac{1}{a+c}$ and $q=\frac{1}{b+d}$. In order to have equality, then $f(x)$ has to have a single root, which is only possible when $b=c$ and the root then will be $x=b=c$ so $f(x)=(p+q)(x-c)^2$. Hence $f(x)=p(x-a)(x-c)+q(x-b)(x-d)=(p+q)(x-c)^2$. Looking at the coefficients we see that $ad=c^2$.