Sequences & Series Question - None

Question

If a, b, c are non-zero real numbers, then the minimum value of the expression

$$\frac{(a^8 + 4a^4 + 1)(b^4 + 3b^2 + 1)(c^2 + 2c + 2)}{a^4b^2}$$

equals

Accepted Answer

Step 1: Rewrite the expression as:

$$P = \left(a^4 + 4 + \frac{1}{a^4} ight)\left(b^2 + 3 + \frac{1}{b^2} ight)\left((c+1)^2 + 1 ight)$$

Step 2: Apply AM-GM inequality to each factor:

$$a^4 + 4 + \frac{1}{a^4} \geq 2\sqrt{a^4 \cdot \frac{1}{a^4}} + 4 = 6$$

$$b^2 + 3 + \frac{1}{b^2} \geq 2\sqrt{b^2 \cdot \frac{1}{b^2}} + 3 = 5$$

$$(c+1)^2 + 1 \geq 1$$

Step 3: Therefore:

$$P \geq 6 imes 5 imes 1 = 30$$

∴ The minimum value is 30.

Question: