Test Index

UPSC CDS I 2024 Math Paper

Question : 60 of 100

Marks: +1, -0

(\frac{a^{2}+3a+1}{a})(\frac{b^{2}+3b+1}{b})

for a, b > 0?

Solution:

Concept:

For positive real numbers

a

and

b

, the minimum value of

a + \frac{1}{a}

2

by AM–GM inequality.

Explanation:

Simplify the given expression:

\left(\frac{a^{2} + 3a + 1}{a}\right) \left(\frac{b^{2} + 3b + 1}{b}\right) = \left(a + 3 + \frac{1}{a}\right) \left(b + 3 + \frac{1}{b}\right)

Since

a > 0

, the minimum of

a + \frac{1}{a}

2

Thus the minimum of the first factor is

2 + 3 = 5

Similarly, the minimum of the second factor is also

5

Hence the minimum value of the product is

5 \times 5 = 25

Answer:

25 (Option D).

Go to Question:

Prev Question

Next Question