Question

Sol. [Key Concept: Four unequal positives in A.P. ⇒ AM–GM gives the inequalities.]

$${\frac{\frac{1}{a} + \frac{1}{\text{ }d}}{\frac{1}{\text{ }b} + \frac{1}{c}} = \frac{\frac{1}{a} + \frac{1}{a + 3{\text{ }d}_{1}}}{\frac{1}{a + d_{1}} + \frac{1}{a + 2{\text{ }d}_{1}}} = \frac{\left( a + d_{1} \right)\left( a + 2{\text{ }d}_{1} \right)}{a\left( a + 3{\text{ }d}_{1} \right)} = 1 + \frac{2{\text{ }d}_{1}^{2}}{a\left( a + 3{\text{ }d}_{1} \right)} > 1 }{ad - bc = a\left( a + 3{\text{ }d}_{1} \right) - \left( a + d_{1} \right)\left( a + 2{\text{ }d}_{1} \right) = - 2{\text{ }d}_{1}^{2} < 0}$$

$$\frac{b + c}{2} > \frac{2}{\frac{1}{b} + \frac{1}{c}} \Rightarrow \frac{1}{b} + \frac{1}{c} > \frac{4}{b + c} = \frac{4}{(a + d)}$$

✔ (a) $\frac{1}{a} + \frac{1}{d} > \frac{1}{b} + \frac{1}{c}$

✘ (b) reverse is false

✔ (c) $ad - bc = - 2d_{1}^{2 < 0} \Rightarrow \ ad < bc$

✔ (d) $\frac{1}{b} + \frac{1}{c} > \frac{4}{b + c} = \frac{4}{a + d}$