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Algebra Level 5

Positive reals a a , b b , and c c are such that a b + b c + a c = 1 ab+bc+ac=1 . Find the minimum value of the following expression:

a 2 b + c + b 2 a + c + c 2 a + b + 36 a + b + c \frac{a^{2}}{b+c}+\frac{b^{2}}{a+c}+\frac{c^{2}}{a+b}+\frac{36}{a+b+c}

Bonus : For what values of a a , b b , and c c , is this minimum achieved?


The answer is 20.

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1 solution

I cannot type latex? I tried typing 1 2 \frac{1}{2} but it displays raw text..

The answer is wrong. It is a value greater than 20. Some number between 20 and 21.

Srikanth Tupurani - 2 years, 1 month ago

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Elaborate please?

Steven Jim - 2 years ago

Try a = 1 α , b = 1 β , c = α + β α β 2 ( α + β ) a=1-\alpha,b=1-\beta,c=\frac{\alpha+\beta-\alpha\beta}{2-(\alpha+\beta)} with α \alpha and β \beta being infinitesimal positive real numbers.

Janardhanan Sivaramakrishnan - 2 years, 1 month ago

Completely wrong answer. try a=b=c=1/√3.It will come greater than 22

Ritabrata Roy - 1 year, 2 months ago

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You have to find the minimum value. Not the maximum.

Janardhanan Sivaramakrishnan - 1 year, 1 month ago

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