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UPSC CDS II 2024 Math Paper

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Question : 33 of 100

Marks: +1, -0

What is the minimum value of

\frac{(a^{8}+a^{4}+1)(b^{8}+b^{4}+1)}{a^{4}b^{4}}

where a > 0, b > 0?

1
4
9
16

Solution:

Concept:

Apply AM‑GM inequality to find the lower bound of the expression.

Explanation:

Let

x = a^4

and

y = b^4

.

Then the expression becomes

\frac{(x^{2}+x+1)(y^{2}+y+1)}{xy}

.

By AM‑GM on

x^{2}

,

x

, and

1

:

\frac{x^{2}+x+1}{3} \ge \sqrt[3]{x^{2}\cdot x\cdot 1}=x

, so

x^{2}+x+1 \ge 3x

.

Similarly,

y^{2}+y+1 \ge 3y

.

Substitute these inequalities:

\frac{(x^{2}+x+1)(y^{2}+y+1)}{xy} \ge \frac{(3x)(3y)}{xy}=9

.

Equality holds when

x=1

and

y=1

, i.e.,

a=b=1

.

Hence the minimum value is

9

.

Answer:

9

(Option C)

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