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Algebra Level 5

Let a , b , c a, b, c be nonnegative real numbers. Minimize

( a − b c ) 2 + ( b − c a ) 2 + ( c − a b ) 2 ( a − b ) 2 + ( b − c ) 2 + ( c − a ) 2 . \large \frac{(a-bc)^2 + (b-ca)^2 + (c-ab)^2}{(a-b)^2 + (b-c)^2 + (c-a)^2}.


The answer is 0.5.

Priyanshu Mishra by Priyanshu Mishra , 20, India

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1 solution

Viki Zeta
Sep 25, 2016

For a=b=c=0, the equation becomes undetermined. Talking a=b=0, and c=1, since a,b,c are integers gives us the minimum value. ⟹ ( a − b c ) 2 + ( b − c a ) 2 + ( c − a b ) 2 ( a − b ) 2 + ( b − c ) 2 + ( c − a ) 2 = ( 0 − ( 0 ) 1 ) 2 + ( ( 0 ) − 1 ( 0 ) ) 2 + ( 1 − ( 0 ) ) 2 ( ( 0 ) − ( 0 ) ) 2 + ( ( 0 ) − 1 ) 2 + ( 1 − 0 ) 2 = 1 1 + 1 = 1 2 = 0.5 \text{For a=b=c=0, the equation becomes undetermined.} \\ \text{Talking a=b=0, and c=1, since a,b,c are integers gives us the minimum value.} \\ \implies \dfrac{(a-bc)^2 + (b-ca)^2 + (c-ab)^2}{(a-b)^2 + (b-c)^2 + (c-a)^2} = \dfrac{(0-(0)1)^2 + ((0)-1(0))^2 + (1-(0))^2}{((0)-(0))^2 + ((0)-1)^2 + (1-0)^2} \\ = \dfrac{1}{1 + 1} \\ = \dfrac{1}{2} = \boxed{ 0.5 }

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Your solution is absolutely wrong.

Harsh Shrivastava - 4 years, 8 months ago

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Then you post one. He just asked the minimum value.

Viki Zeta - 4 years, 8 months ago

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I am trying the problem but have not got it.

Harsh Shrivastava - 4 years, 8 months ago

@Vicky Vignesh - Yes. And your solution said integers. I was wondering why you decided to disclude so many numbers.

Emily Namm - 4 years, 8 months ago

The problem does not say a,b,c are integers

Emily Namm - 4 years, 8 months ago

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But it said they are non-negative real numbers

Viki Zeta - 4 years, 8 months ago

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