Maxima of Sum of Reciprocals!

Number Theory Level 4

1 a + 1 b + 1 c < 1 \large{\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} < 1}

If a , b , c a,b,c are three positive integers satisfying the above inequality, then let S S be the maximum possible value of 1 a + 1 b + 1 c \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} . If S S can be expressed as P Q \dfrac{P}{Q} for positive coprime integers P , Q P,Q then find the value of P + Q P+Q .


The answer is 83.

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Satyajit Mohanty by Satyajit Mohanty , 24, India

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1 solution

Karan Arora
Sep 12, 2015

We know that a,b,c are positive integers. 1 a + 1 b + 1 c \frac{1}{a} + \frac{1}{b} + \frac{1}{c} <1 As we know for f(x)= 1 x \frac{1}{x} for int As x increases f(x) decreases so we have to choose 3 maximum values for S to be maximum. 1/2 is the maximum value for f(x) where x>1 Then,1/3,1/4 and so on.

Firstly adding the three maximum value i.e. 1 2 + 1 3 + 1 3 \frac{1}{2} + \frac{1}{3} + \frac{1}{3} = 13 12 \frac{13}{12} >1 Similarly for 1/2,1/3,1/5 => S>1& 1/2,1/3,1/6 => S=1 Hence for 1/2,1/3,1/7 S will be maximum 1 2 + 1 3 + 1 7 = 41 42 \frac{1}{2} + \frac{1}{3} + \frac{1}{7} =\frac{41}{42} P+Q=41+42= 83 \boxed{83}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

This is not correct. Why must we choose the maximum of the two numbers (2,3) first? Even after "WLOG, a ≤ b ≤ c a\leq b\leq c ", why can't we start with a = 3 a = 3 ? You didn't justify that a = 2 a=2 is only possible.

Pi Han Goh - 5 years, 6 months ago

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