Question

Let a₁, a₂, a₃,.....,a_n be positive real numbers.Then the minimum value of \(\frac{a_1}{a_2}+\frac{a_2}{a_3}+....+\frac{a_n}{a_1}\) is

• A. 1
• B. n
• C. ⁿC₂
• D. 2

Answer 1

The Correct Option is B

Step 1: Given positive real numbers $a_1, a_2, \dots, a_n$ such that:
$a_1 \cdot a_2 \cdot \dots \cdot a_n = c$ 

Step 2: We want to minimize the expression: 
$a_1 + a_2 + \dots + a_{n-1} + 2a_n$ 

Step 3: By the AM ≥ GM inequality, for any positive real numbers:
$\displaystyle \frac{a_1 + a_2 + \dots + a_{n-1} + 2a_n}{n} \geq \left(a_1 a_2 \dots a_{n-1} (2a_n)\right)^{1/n}$ 

Step 4: Using the given condition:
$a_1 a_2 \dots a_n = c \Rightarrow a_1 a_2 \dots a_{n-1} = \frac{c}{a_n}$ 

So:
$\displaystyle \left(a_1 a_2 \dots a_{n-1} \cdot 2a_n\right) = \frac{c}{a_n} \cdot 2a_n = 2c$ 

Step 5: Plug into the inequality:
$\displaystyle \frac{a_1 + a_2 + \dots + a_{n-1} + 2a_n}{n} \geq (2c)^{1/n}$ 

Step 6: Multiply both sides by $n$:
$a_1 + a_2 + \dots + a_{n-1} + 2a_n \geq n(2c)^{1/n}$ 

Step 7: The minimum value is therefore:
$\displaystyle \min(a_1 + a_2 + \dots + a_{n-1} + 2a_n) = n(2c)^{1/n}$ 

Step 8: If $c = \left(\frac{1}{2}\right)^n$, then:
$\displaystyle n(2c)^{1/n} = n \left(2 \cdot \left(\frac{1}{2}\right)^n \right)^{1/n} = n \left(2^{1 - n} \right)^{1/n} = n \cdot 2^{(1 - n)/n}$ If $c = \frac{1}{2^n}$, then this simplifies to $n$. 

Final Answer: (B): $n$

Answer 2

Let \( a_1, a_2, a_3, \dots, a_n \) be positive real numbers. We want to find the minimum value of the expression:

\( \displaystyle \frac{a_1}{a_2} + \frac{a_2}{a_3} + \dots + \frac{a_{n-1}}{a_n} + \frac{a_n}{a_1} \)

We apply the AM-GM inequality to the \( n \) terms: 

\( \displaystyle \frac{\frac{a_1}{a_2} + \frac{a_2}{a_3} + \dots + \frac{a_{n-1}}{a_n} + \frac{a_n}{a_1}}{n} \geq \sqrt[n]{\frac{a_1}{a_2} \cdot \frac{a_2}{a_3} \cdot \dots \cdot \frac{a_{n-1}}{a_n} \cdot \frac{a_n}{a_1}} \)

The product inside the \( n \)-th root simplifies to:

\( \displaystyle \frac{a_1}{a_2} \cdot \frac{a_2}{a_3} \cdot \dots \cdot \frac{a_{n-1}}{a_n} \cdot \frac{a_n}{a_1} = 1 \)

So the inequality becomes:

\( \displaystyle \frac{\frac{a_1}{a_2} + \frac{a_2}{a_3} + \dots + \frac{a_{n-1}}{a_n} + \frac{a_n}{a_1}}{n} \geq \sqrt[n]{1} = 1 \)

Multiplying both sides by \( n \), we get:

\( \displaystyle \frac{a_1}{a_2} + \frac{a_2}{a_3} + \dots + \frac{a_{n-1}}{a_n} + \frac{a_n}{a_1} \geq n \)

The minimum value is achieved when all the terms are equal, i.e., \( \frac{a_1}{a_2} = \frac{a_2}{a_3} = \dots = \frac{a_n}{a_1} \), which happens when \( a_1 = a_2 = \dots = a_n \). In this case, each term becomes 1, and the sum is \( n \).

Therefore, the minimum value of the expression is \( n \).