(i) If $a, b, c, d$ are four distinct positive quantities in A.P., then show that $b c>a d$ (ii) If $a, b, c, d$ are four distinct positive quantities in G.P., then show that $$a+d>b+c$$ $\textbf{Solution}$ (i) Since $a, b, c, d$ are in A.P., then A.M. > G.M., for the first three terms. Therefore, $b>\sqrt{a c}$ Squaring, we get $$\left(\text { Here } \frac{a+c}{2}=b\right)$$ $b^{2}>a c$ Similarly, for the last three terms $\mathrm{AM}>\mathrm{GM}$ $c>\sqrt{b d}$ $$\left(\text { Here } \frac{b+d}{2}=c\right)$$ $c^{2}>b d$ Multiplying (1) and (2), we get $$\begin{aligned} & b^{2} c^{2}>(a c)(b d) \\ \Rightarrow & b c>a d \\ \text { (ii) } & \text { Since } a, b, c, d \text { are in G.P. } \end{aligned}$$ again A.M. > G.M. for the first three terms $$\begin{aligned} & \frac{a+c}{2}>b \quad(\text { since } \sqrt{a c}=b) \\ \Rightarrow & a+c>2 b \end{aligned}$$ Similarly, for the last three terms $$\begin{aligned} & \frac{b+d}{2}>c \\ \Rightarrow & b+d>2 c \end{aligned}\left(\begin{array}{l} \text { since } \sqrt{b d}=c) \\ \end{array}\right.$$ Adding (3) and (4), we get $$\begin{aligned} & (a+c)+(b+d)>2 b+2 c \\ & a+d>b+c \end{aligned}$$ Eample 12 If $a, b, c$ are three consecutive terms of an A.P. and $x, y, z$ are three consecutive terms of a G.P. Then prove that $$x^{b-c} \cdot y^{c-a} \cdot z^{a-b}=1$$

Step 1: Analyze the properties of A.P. for part (i)

For part (i), we are given that a, b, c, d are four distinct positive quantities in an Arithmetic Progression (A.P.). In an A.P., the middle term is the arithmetic mean of its neighbors. Thus, $$b$$ is the arithmetic mean of $$a$$ and $$c$$ , and $$c$$ is the arithmetic mean of $$b$$ and $$d$$ .

Step 2: Apply AM-GM inequality for part (i)

Since a, b, c are distinct positive quantities, the Arithmetic Mean (AM) is strictly greater than the Geometric Mean (GM). So, $$b > \sqrt{ac}$$ , which implies $$b^2 > ac$$ . Similarly, for b, c, d, we have $$c > \sqrt{bd}$$ , which implies $$c^2 > bd$$ .

Step 3: Multiply the inequalities for part (i)

Multiplying the two inequalities $$b^2 > ac$$ and $$c^2 > bd$$ , we get $$b^2 c^2 > (ac)(bd)$$ . Since all quantities are positive, we can take the square root of both sides, which simplifies to bc > ad. This proves the first part.

Step 4: Analyze the properties of G.P. for part (ii)

For part (ii), we are given that a, b, c, d are four distinct positive quantities in a Geometric Progression (G.P.). In a G.P., the middle term is the geometric mean of its neighbors. Thus, $$b$$ is the geometric mean of $$a$$ and $$c$$ , and $$c$$ is the geometric mean of $$b$$ and $$d$$ .

Step 5: Apply AM-GM inequality for part (ii)

Since a, b, c are distinct positive quantities, the Arithmetic Mean (AM) is strictly greater than the Geometric Mean (GM). So, $$\frac{a+c}{2} > \sqrt{ac}$$ . Since $$b = \sqrt{ac}$$ , we have $$\frac{a+c}{2} > b$$ , which implies a+c > 2b. Similarly, for b, c, d, we have $$\frac{b+d}{2} > \sqrt{bd}$$ . Since $$c = \sqrt{bd}$$ , we get $$\frac{b+d}{2} > c$$ , which implies b+d > 2c.

Step 6: Add the inequalities for part (ii)

Adding the two inequalities a+c > 2b and b+d > 2c, we get (a+c) + (b+d) > $2b + 2c$. Rearranging the terms, we subtract $$b$$ and $$c$$ from both sides, which simplifies to a+d > b+c. This proves the second part.

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