Sequences & SeriesmediumFree

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)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Sequences & Series · easy
a, b, c, d are in increasing G.P. If the A.M. between a and b is 6 and the A.M. between c…
Sequences & Series · easy
Let a₁, a₂, … and b₁, b₂, … be arithmetic progressions with a₁=25 , b₁=75 and a(100)+b(10…
Sequences & Series · easy
If S_n=(1)/(1·2·3)+(1)/(2·3·4)+⋯+(1)/(n(n+1)(n+2)) , then 8S_∞ is
Sequences & Series · easy
If the sum to infinity of 1+4x+7x²+10x³+⋯ is (35)/(16) , where |x|<1 , then x equals

Solve more, learn faster

Sign up free to solve more JEE Mathsquestions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath