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AM-GM Inequality Minimum Value

XAT 2013 · Quantitative Ability & Data Interpretation · Medium · Algebra

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Consider the expression (a2+a+1)(b2+b+1)(c2+c+1)(d2+d+1)(e2+e+1)abcde\frac{(a^2+a+1)(b^2+b+1)(c^2+c+1)(d^2+d+1)(e^2+e+1)}{abcde}, where a,b,c,d and e are positive numbers. The minimum value of the expression is

  1. A.

    3

  2. B.

    1

  3. C.

    10

  4. D.

    100

  5. E.

    243

Answer

E

Explanation

Rewrite as (a+1+1a)(b+1+1b)(c+1+1c)(d+1+1d)(e+1+1e)\left(a + 1 + \frac{1}{a}\right)\left(b + 1 + \frac{1}{b}\right)\left(c + 1 + \frac{1}{c}\right)\left(d + 1 + \frac{1}{d}\right)\left(e + 1 + \frac{1}{e}\right). By AM-GM, for any positive xx, x+1x≥2  ⟹  x+1+1x≥3x + \frac{1}{x} \ge 2 \implies x + 1 + \frac{1}{x} \ge 3. Minimum value = 35=2433^5 = 243.

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