An algebra problem by Anish Roy
True or False?
If a , b , and c are positive real numbers such that a + 1 a + b + 1 b + c + 1 c = 1 , then the maximum value of a b c is 8 1 .
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2 solutions
i did it by AM -GM -HM
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Please post your solution! I bet it would be more elegant than old school calculus.
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how can i upload a solution(like i solved it in my notebook ,but dont know how to upload its picture. ) ? actually i ticked false , so i am not having an option to upload my answer
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@Sam Dave – Just take a photo and upload it here. Or, if you have time, try to learn the LaTeX.
This is a very usual try, using calculus to solve any question regarding maximum and minimum. However, can you do it via AM-GM-HM? I can't Moreover, how do you know that it is maximum and not minimum?
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You use the Second Partials Test to show that it is a maximum and not a minimum:
∂ x ∂ ∂ x ∂ = − ( 2 x y + x + y ) 3 2 x 2 ( y + 1 ) 2 ∂ y ∂ ∂ y ∂ = − ( 2 x y + x + y ) 3 2 y 2 ( x + 1 ) 2 ∂ y ∂ ∂ x ∂ = − ( 2 x y + x + y ) 3 2 x y ( x + 1 ) ( y + 1 ) ( 2 x y + x + y − 1 )
Evaluating each of these at x , y = 2 1 , we find that f x x ( 0 . 5 , 0 . 5 ) = f y y ( 0 . 5 , 0 . 5 ) = − 3 1 f x y ( 0 . 5 , 0 . 5 ) = − 6 1 leads to f x x ( 0 . 5 , 0 . 5 ) ∗ f y y ( 0 . 5 , 0 . 5 ) − ( f x y ( 0 . 5 , 0 . 5 ) ) 2 = − 3 1 ∗ ( − 3 1 ) − ( − 6 1 ) 2 = 9 1 − 3 6 1 = 1 2 1 > 0
⇒ f ( 0 . 5 , 0 . 5 ) is not a saddle point .
Therefore, since f x x ( 0 . 5 , 0 . 5 ) = − 3 1 < 0 , ( 0 . 5 , 0 . 5 ) must be a local maximum of f ( x , y ) .
As for whether there exists a quick and clever way to use inequalities for this problem, you should ask @Anish Roy .
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Thanks! My calculus has not reached that level though :( I just assumed that a, b, and c are all equal and try if they are not. And I get that if they are not equal, then the value is less than 1/8. Lol.
This is equivalent to ∑ a ( 1 + b ) ( 1 + c ) = ( 1 + a ) ( 1 + b ) ( 1 + c ) . This simplifies to a b + b c + c a + 2 a b c = 1 Using AM-GM inequality, we have 1 = a b + b c + c a + 2 a b c ≥ 4 ( a b ⋅ b c ⋅ c a ⋅ 2 a b c ) 1 / 4 Simplificaton gives a b c ≤ 8 1
Manipulation yields c = 2 a b + a + b 1 − a b . Thus a b c = 2 a b + a + b ( a b ) ( 1 − a b ) .
After this substitution, the question can now be interpreted as asking for the maximum value of f ( x , y ) = 2 x y + x + y x y ( 1 − x y ) over real x , y > 0 .
Clearly, m a x ( f ( x , y ) ) will occur at the point where ∂ x ∂ = 0 and ∂ y ∂ = 0 . To find this point, we solve ∂ x ∂ = ( 2 x y + x + y ) 2 − y 2 ( x + 1 ) ( x + 2 x y − 1 ) ∂ y ∂ = ( 2 x y + x + y ) 2 − x 2 ( y + 1 ) ( y + 2 x y − 1 )
Notice now since x , y > 0 , we must have 2 x y + x − 1 = 0 and 2 x y + y − 1 = 0 .
Therefore, x = y as a result of 2 x y + x − 1 − ( 2 x y + y − 1 ) = 0 − ( 0 ) ⇒ x − y = 0 , meaning that we can solve directly 2 x 2 + x − 1 = 0 → x = 2 1 , − 1 , with − 1 < 0 rejected, leaves only x = 0 . 5 .
Hence, calculating the maximum value of a b c = 2 x y + x + y ( x y ) ( 1 − x y ) = 2 ∗ 0 . 5 ∗ 0 . 5 + 0 . 5 ∗ 0 . 5 0 . 5 ∗ 0 . 5 ∗ ( 1 − 0 . 5 ∗ 0 . 5 ) = 8 1 . The statement is T R U E .