Question

If $p,q$ are positive real numbers such that $pq = 1$ , then the least value of $(1 + p) (1 + q)$ is

• A. 4
• B. 1
• C. 2
• D. None of these

Answer 1

Answer: (A)

4

Answer 2

The correct option is(A): 4.

From the information given, we have:

1. ap =1, which implies a =1 (since a and p are positive, and any non-zero number raised to the power of 0 is 1).
2. ab =1, which implies b =a 1​=1.
3. pq =1, which means p =q 1​ and =p 1​.

Now let's focus on the expression to be minimized:

(1+a)(1+b)(1+p)(1+q)

Substitute the values of a , b , p , and q :

(1+1)(1+1)(1+q 1​)(1+p 1​)

Simplify:

(2)(2)(1+q 1​)(1+p 1​)

(4)(1+q 1​)(1+p 1​)

Now, use the information that p and q are positive real numbers and pq =1:

By the AM-GM inequality (Arithmetic Mean-Geometric Mean inequality), for any two positive real numbers x and y , we have:

2 x +y ​≥ xy ​

For x =p and y =q , we have:

2 p +q ​≥ pq ​

Since pq =1, this simplifies to:

p +q/2 ​≥1

Now, let's apply this to the expression we want to minimize:

(4)(1+q 1​)(1+p 1​)=4⋅21+q ​⋅21+p ​

By the AM-GM inequality:

21+q ​≥1⋅ q ​=q ​

21+p ​≥1⋅ p ​=p ​

Multiply these inequalities:

1+q ​⋅21+p ​≥ pq ​=1

So,

1=44⋅21+q ​⋅21+p ​≥4⋅1=4

Hence, the least value of (1+a)(1+b)(1+p)(1+q) is 4.