Harmonic progression based question

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?

Answer 1

The correct option is $(C)$.

We have $$\frac 1a+\frac 1c=\frac 2b\quad\text{and}\quad \frac 1b+\frac 1d=\frac 2c$$$$\Rightarrow c=\frac{ab}{2a-b}\quad\text{and}\quad d=\frac{ab}{3a-2b}$$ Here, since $c,d$ are positive, we have $2a-b\gt 0,3a-2b\gt 0$.

By the way, we have $$\begin{align}a+d-(b+c)&=a+\frac{ab}{3a-2b}-b-\frac{ab}{2a-b}\\&=\frac{(a-b)(3a-2b)(2a-b)+ab(2a-b)-ab(3a-2b)}{(3a-2b)(2a-b)}\\&=\frac{(a-b)(3a-2b)(2a-b)-ab(a-b)}{(3a-2b)(2a-b)}\\&=\frac{(a-b)((3a-2b)(2a-b)-ab)}{(3a-2b)(2a-b)}\\&=\frac{(a-b)\cdot 2(3a^2-4ab+b^2)}{(3a-2b)(2a-b)}\\&=\frac{2 (a-b)^2 (3 a-b)}{(3a-2b)(2a-b)}\end{align}$$ This is positive because $3a-2b\gt 0\Rightarrow 3a-b\gt 0$.

Hence, $a+d\gt b+c$ holds.

(By the way, options $(A),(B)$ are wrong. Take $a=2,b=5/2$. So, the correct option is $(C)$ only.)

Answer 2

$$a=\dfrac1{x-3y},b=\dfrac1{x-y},c=\dfrac1{x+y},d=\dfrac1{x+3y}$$

$\implies a+d=\dfrac{2x}{x^2-9y^2}$ and $b+c=\dfrac{2x}{x^2-y^2}$

$\implies a+d>b+c$ as $9y^2>y^2$