Harmonic progression based question 
If $a,b,c,d$ are distinct positive numbers in harmonic progression, then
(A) $a+b>c+d$
(B) $a+c>b+d$
(C) $a+d>b+c$
(D) none of these

I tried $\frac{1}{a},\frac{1}{b},\frac{1}{c},\frac{1}{d}$ are in AP so either $\frac{1}{a}>\frac{1}{b}>\frac{1}{c}>\frac{1}{d}$ or $\frac{1}{a}<\frac{1}{b}<\frac{1}{c}<\frac{1}{d}$.That means either $a>b>c>d$ or $a<b<c<d$.But i could not judge correct answer based on this.Is my approach not correct?What is correct answer?Can someone help me in solving this question?

$$a=\dfrac1{x-3y},b=\dfrac1{x-y},c=\dfrac1{x+y},d=\dfrac1{x+3y}$$
$\implies a+d=\dfrac{2x}{x^2-9y^2}$ and $b+c=\dfrac{2x}{x^2-y^2}$
$\implies a+d>b+c$ as $9y^2>y^2$
