It is not even cyclic!

Algebra Level 5

Positive real a , b , a, \space b, and c c are such that a + b + c = 1 a+b+c=1 . Find the minimum value of a + b a b c \dfrac{a+b}{abc} .

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The answer is 16.

Fidel Simanjuntak by Fidel Simanjuntak , 19, Indonesia

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1 solution

a + b a b c ≥ 2 a b c Since a + b 2 ≥ a b ≥ 4 ( a + b ) c Since a b ≤ a + b 2 = 4 c ( 1 − c ) ( a + b = 1 − c ) ≥ 16 Since c ( 1 − c ) ≤ ( c + 1 − c 2 ) 2 = 1 4 \displaystyle \begin{aligned} \dfrac{a+b} {abc} &\ge \dfrac{2}{\sqrt{ab}c}\quad \text{Since }\dfrac{a+b} {2}\ge \sqrt{ab} \\ &\ge \dfrac{4}{(a+b)c}\quad\text{Since } \sqrt{ab} \le \dfrac{a+b}{2} \\ & = \dfrac{4}{c(1-c)}\quad \left(a+b=1-c\right) \\ & \ge 16\quad \text{Since } c(1-c)\le \left(\dfrac{c+1-c}{2}\right) ^2=\dfrac{1}{4} \end{aligned}

Equality holds when a = b = 0.25 , c = 0.5 a=b=0.25,c=0.5

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You need to state that the value of 16 16 can be achieved, when c = 1 2 c=\tfrac12 and a = b = 1 4 a=b=\tfrac14 .

Mark Hennings - 4 years, 1 month ago

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Sure i will add that in, I forgot to add the equality cases

Aditya Narayan Sharma - 4 years, 1 month ago

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