Problem 1

Algebra Level 5

x 2 y + 2 z + y 2 z + 2 x + z 2 x + 2 y \frac{x^2}{y+2z}+\frac{y^2}{z+2x}+\frac{z^2}{x+2y}

Given that x x , y y and z z are positive reals which satisfy 3 ( x 4 + y 4 + z 4 ) − 7 ( x 2 + y 2 + z 2 ) = − 12 3(x^4+y^4+z^4)-7(x^2+y^2+z^2)=-12 . Let the minimum value of the expression above be P P . Find ⌊ 10000 P ⌋ \left \lfloor 10000P \right \rfloor .

This problem is in this set .


The answer is 10000.

Son Nguyen by Son Nguyen , 20, Vietnam

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1 solution

Sal Gard
May 4, 2016

The given conditions are symmetric, which means a minimum will be attained when x=y=z. Since the following equation holds, it can be turned into a biquadratic with solutions 1 and 2sqrt(3)/3. Trying out each value, it is evident that the expression is the same as the value so therefore, the minimum is 1 and the answer is 10000.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

How do we prove that minimum is attained for x=y=z.

Sahil Jain - 3 years, 7 months ago

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