Question

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

Accepted Answer

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

AM-GM Inequality Minimum Value

Explanation

Practise this under exam conditions