Question

If the product of three positive real numbers say \(a,b,c\) be \(27\), then the minimum value of \(ab+bc+ca\) is equal to

• A. \(27^{4/3}\)
• B. \(27^{1/3}\)
• C. \(27^{2/3}\)
• D. \(27\)

Hint:

When using AM-GM on \(ab+bc+ca\), remember to multiply by \(3\) after applying: \[ \frac{ab+bc+ca}{3}\geq \sqrt[3]{(ab)(bc)(ca)} \]

Answer 1

The Correct Option is D

Concept:
This question is based on AM-GM inequality. For positive real numbers \(x,y,z\): \[ \frac{x+y+z}{3}\geq \sqrt[3]{xyz} \] Here we need the minimum value of: \[ ab+bc+ca \] Given: \[ abc=27 \]

Step 1: Apply AM-GM to \(ab\), \(bc\), and \(ca\).
Since \(a,b,c\) are positive real numbers, \(ab\), \(bc\), and \(ca\) are also positive. So: \[ \frac{ab+bc+ca}{3}\geq \sqrt[3]{(ab)(bc)(ca)} \]

Step 2: Simplify the expression inside cube root.
\[ (ab)(bc)(ca)=a^2b^2c^2 \] \[ a^2b^2c^2=(abc)^2 \] So: \[ \sqrt[3]{(ab)(bc)(ca)}=\sqrt[3]{(abc)^2} \] \[ =(abc)^{2/3} \]

Step 3: Substitute \(abc=27\).
\[ \frac{ab+bc+ca}{3}\geq 27^{2/3} \] Now: \[ 27^{1/3}=3 \] So: \[ 27^{2/3}=3^2=9 \] Therefore: \[ \frac{ab+bc+ca}{3}\geq 9 \] \[ ab+bc+ca\geq 27 \]

Step 4: Check equality condition.
Equality in AM-GM occurs when: \[ ab=bc=ca \] Since \(a,b,c>0\), this gives: \[ a=b=c \] Given: \[ abc=27 \] So: \[ a=b=c=3 \] Then: \[ ab+bc+ca=9+9+9=27 \] So the minimum value is: \[ 27 \]

Step 5: Check the options.
Option (A) \(27^{4/3}\) is not correct.
Option (B) \(27^{1/3}=3\) is too small.
Option (C) \(27^{2/3}=9\) is the AM-GM lower value before multiplying by \(3\), so it is not the final minimum.
Option (D) \(27\) is correct. Hence, the correct answer is: \[ \boxed{(D)\ 27} \]