<s>[INST] 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?
 [/INST] 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.)
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