Question

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

• A. 3
• B. 1
• C. 10
• D. 100
• E. 243

Hint:

Use the identity \(\dfracx^2+x+1x=x+1+\dfrac1x\) and the basic AM–GM fact \(x+\dfrac1x\ge2\) to get simple lower bounds. For products of identical lower-bounded factors, raise the bound to the appropriate power.

Answer 1

Answer: (E)

Step 1: Rewrite the expression by separating each factor:
\[ \frac{(a^2+a+1)(b^2+b+1)(c^2+c+1)(d^2+d+1)(e^2+e+1)}{abcde} =\prod_{x\in\{a,b,c,d,e\}}\frac{x^2+x+1}{x}. \]

Step 2: For a positive variable \(x\),
\[ \frac{x^2+x+1}{x}=x+1+\frac{1}{x}. \]
By AM–GM (or the inequality \(x+\dfrac{1}{x}\ge 2\) for \(x>0\)),
\[ x+\frac{1}{x}\ge 2 \; \Rightarrow \; x+1+\frac{1}{x}\ge 3, \]
with equality iff \(x=1\).

Step 3: Applying this bound to each factor and multiplying,
\[ \prod_{x\in\{a,b,c,d,e\}}\Big(x+1+\frac{1}{x}\Big)\ge 3^5=243. \]
Equality occurs when \(a=b=c=d=e=1\).

Conclusion: The minimum value of the given expression is 243.