Descartes' Rule of Signs

After some re-arrangements, we have $\displaystyle (b-a)(a+b)+(d-c)(c+d)=2(ac-bd)$ $\displaystyle \iff (b+d)^2=(a+c)^2\iff b+d=\pm(a+c)$

Now notice that all the roots of $f(x)=x^4+ax^3+bx^2+cx+d$ are positive. Hence, according to the Descartes' Rule of Signs , we have that $a,c$ are negative and $b,d$ are positive (since the leading coefficient is positive, and the signs have to alternate, we have that $a$ is negative, $b$ is positive, $c$ is negative and $d$ is positive).

This leads us to the conclusion that $\displaystyle b+d=-(a+c)\iff a+b+c+d=0$ $\displaystyle\iff (a+b+c+d)^2=\boxed{0}$