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Minimum of (a+1/b)^2+(b+1/c)^2+(c+1/a)^2, a+b+c=6 | JEE

JEE Maths question with a full step-by-step solution.

Question
If a,b,ca, b, c are positive with a+b+c=6a+b+c=6, then the minimum value of (a+1b)2+(b+1c)2+(c+1a)2\left(a+\frac1b\right)^2+\left(b+\frac1c\right)^2+\left(c+\frac1a\right)^2 is
A752\frac{75}{2}
B754\frac{75}{4}correct
C654\frac{65}{4}
D652\frac{65}{2}
Solution
Step 1: AM–HM on 1a,1b,1c\frac1a,\frac1b,\frac1c:
1a+1b+1c3≥3a+b+c=12 ⇒ 1a+1b+1c≥32.\frac{\frac1a+\frac1b+\frac1c}{3}\ge\frac{3}{a+b+c}=\frac12\ \Rightarrow\ \frac1a+\frac1b+\frac1c\ge\frac32.
 Step 2: Mean of squares ≥\ge square of mean, applied to the three brackets:
(a+1b)2+(b+1c)2+(c+1a)23≥((a+b+c)+(1a+1b+1c)3)2.\frac{\left(a+\frac1b\right)^2+\left(b+\frac1c\right)^2+\left(c+\frac1a\right)^2}{3}\ge\left(\frac{(a+b+c)+\left(\frac1a+\frac1b+\frac1c\right)}{3}\right)^2.
 Step 3: The inner bracket is ≥6+323=52\ge\dfrac{6+\frac32}{3}=\dfrac52, so
sum≥3(52)2=754.\text{sum}\ge 3\left(\frac52\right)^2=\frac{75}{4}.
 Correct answer: (2)
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