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Algebra Level 4

4 81 ( a b + b c + c a ) + a b c \dfrac{4}{81(ab+bc+ca)}+abc

Given that a , b a,b and c c are positive reals and a + b + c = 1 a+b+c=1 .

If the minimum value of the expression above can be expressed as m n \dfrac{m}{n} where m m and n n are coprime positive integers, find m + n m+n .


The answer is 32.

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1 solution

Son Nguyen
Feb 19, 2016

Using Schur inequality: ( a + b + c ) 3 + 9 a b c 4 ( a + b + c ) ( a b + b c + c a ) (a+b+c)^3+9abc\geq 4(a+b+c)(ab+bc+ca) a b c 4 ( a b + b c + c a ) 1 9 = 4 ( a b + b c + c a ) 9 1 9 abc\geq \frac{4(ab+bc+ca)-1}{9}=\frac{4(ab+bc+ca)}{9}-\frac{1}{9} So we get: 4 81 ( a b + b c + c a ) + 4 ( a b + b c + c a ) 9 1 9 \frac{4}{81(ab+bc+ca)}+\frac{4(ab+bc+ca)}{9}-\frac{1}{9} By applying AM-GM inequality we get: 4 81 ( a b + b c + c a ) + 4 ( a b + b c + c a ) 9 1 9 2 16 729 1 9 = 5 27 = m n \frac{4}{81(ab+bc+ca)}+\frac{4(ab+bc+ca)}{9}-\frac{1}{9}\geq 2\sqrt{\frac{16}{729}}-\frac{1}{9}=\frac{5}{27}=\frac{m}{n} The equality holds when a = b = c = 1 3 a=b=c=\frac{1}{3}

So m + n = 32 \large m+n=32

Can you post more Schur's inequality problems? It's really hard to find one here.

Pi Han Goh - 5 years, 3 months ago

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Yeap.I have a lot of Schur problem...

Son Nguyen - 5 years, 3 months ago

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How did you arrived the name of this problem?

Department 8 - 5 years, 3 months ago

the problem was so interesting

fahim saikat - 5 years, 3 months ago

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