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Solving equality with inequalities: TIFR GS 2017, Part 1 Problem 2

Understand the problem

If \((1+a+b+c)(1+\frac 1a+\frac 1b+\frac 1c)=16\)

then is it true that \(a+b+c=3\)?

Source of the problem
TIFR GS 2017, Part 1 Problem 2

Topic
Inequality
Difficulty Level
Easy
Suggested Book

Introduction to Inequalities (New Mathematical Library)

Start with hints

Do you really need a hint? Try it first!

Try to use AM-GM-HM inequalities
We know \(A.M \geq H.M\). Can you use that?
\(A.M \geq H.M\) when are they equal?
Taking \(1,a,b,c\) we know \(A.M \geq H.M \) then   \(\frac{1+a+b+c}{4}\geq \frac{4}{1+\frac 1a+\frac 1b+\frac 1c} \Rightarrow (1+a+b+c)(1+\frac 1a+\frac 1b+\frac 1c)\geq 16 \).

Now they are equal when \(a=b=c=1\) i.e \(a+b+c=3\).

Hence the statement is true.

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