What is the <strong>MINIMUM</strong> value of \frac{(a^8+a^4+1)(b^8+b^4+1)}{a^4b^4}, where a \gt 0, b \gt 0?

Explanation

Rearrange as (a^4+1+\frac{1}{a^4})(b^4+1+\frac{1}{b^4}). Since x+\frac{1}{x} \geq 2 for x \gt 0, the minimum for a^4+\frac{1}{a^4} is 2. Thus, the minimum value is (2+1)(2+1) = 3 \times 3 = 9.

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