Power Inequality

Algebra Level 3

Let a , b a,b and c c be positive numbers such that a + b + c = 1 a+b+c = 1 .

If the maximum value of a b b c c a a^b b^c c^a can be expressed as m n \dfrac mn , where m m and n n are coprime positive integers, find m + n m+n .


The answer is 4.

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1 solution

Reynan Henry
Dec 26, 2016

with AM-GM 1 = ( a + b + c ) 2 3 ( a b + b c + c a ) 1=(a+b+c)^2\ge 3(ab+bc+ca) so 1 3 a b + b c + c a \frac{1}{3}\ge ab+bc+ca . On the other hand a b + b c + c a = a b + b c + c a a + b + c a b b c c a a + b + c = a b b c c a ab+bc+ca = \frac{ab+bc+ca}{a+b+c}\ge \sqrt[a+b+c]{a^bb^cc^a}=a^bb^cc^a (this is called weighted AM-GM) so the answer is 1 3 \frac{1}{3}

How ab + BC+ca = that thing

ankit raj - 4 years, 5 months ago

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because a + b + c = 1 a+b+c = 1 and dividing something with one is not changing anything

Reynan Henry - 4 years, 5 months ago

For completeness, you should show the equality case.

Sharky Kesa - 4 years, 5 months ago

hey, how did you bring out the first inequality by directly using AM-GM? I can't figure it out.

Dhrubajyoti Ghosh - 3 years, 10 months ago

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