Remarkable Question

Algebra Level 3

1 a + b c + 1 b + a c + 1 c + a b = x ( 1 + a b ) ( 1 + b c ) ( 1 + c a ) \large \dfrac1{a+bc} +\dfrac1{b+ac} + \dfrac1{c+ab} = \dfrac x{ \left(1 + \frac ab\right)\left(1 + \frac bc\right)\left(1 + \frac ca\right) }

Let a , b a,b and c c be positive real numbers satisfying 1 a + 1 b + 1 c = 1 \dfrac1a + \dfrac1b + \dfrac1c = 1 , find the value of x x satisfying the algebraic identity above.


The answer is 2.

حافظ القران الكريم by حافظ القران الكريم , 23, Saudi Arabia

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2 solutions

Hung Woei Neoh
May 13, 2016

1 a + b c + 1 b + a c + 1 c + a b = x ( 1 + a b ) ( 1 + b c ) ( 1 + c a ) \dfrac{1}{a+bc} + \dfrac{1}{b+ac} + \dfrac{1}{c+ab} = \dfrac{x}{\left(1+\dfrac{a}{b} \right) \left(1+\dfrac{b}{c} \right) \left(1+\dfrac{c}{a} \right)}\\

Given that 1 a + 1 b + 1 c = 1 ⟹ \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = 1 \implies Eq.(1)

b c + a c + a b a b c = 1 \dfrac{bc+ac+ab}{abc}=1

a b + a c + b c = a b c ⟹ ab+ac+bc=abc \implies Eq.(2)

We will use these two equations to manipulate the identity.

Use Eq.(2) on the LHS:

1 a + b c + 1 b + a c + 1 c + a b = 1 a + a b c − a b − a c + 1 b + a b c − a b − b c + 1 c + a b c − a c − b c = 1 a ( b c − b − c + 1 ) + 1 b ( a c − a − c + 1 ) + 1 c ( a b − a − b + 1 ) = 1 a ( b − 1 ) ( c − 1 ) + 1 b ( a − 1 ) ( c − 1 ) + 1 c ( a − 1 ) ( b − 1 ) \dfrac{1}{a+bc} + \dfrac{1}{b+ac} + \dfrac{1}{c+ab}\\ =\dfrac{1}{a+abc-ab-ac} + \dfrac{1}{b+abc-ab-bc} + \dfrac{1}{c+abc-ac-bc}\\ =\dfrac{1}{a(bc-b-c+1)} + \dfrac{1}{b(ac-a-c+1)} + \dfrac{1}{c(ab - a- b+1)}\\ =\dfrac{1}{a(b-1)(c-1)} + \dfrac{1}{b(a-1)(c-1)} + \dfrac{1}{c(a-1)(b-1)}

Next, we use Eq.(1) on the RHS:

x ( 1 + a b ) ( 1 + b c ) ( 1 + c a ) = x a ( 1 a + 1 b ) b ( 1 b + 1 c ) c ( 1 c + 1 a ) = x a b c ( 1 − 1 c ) ( 1 − 1 a ) ( 1 − 1 b ) = x a b c ( c − 1 c ) ( a − 1 a ) ( b − 1 b ) = x a b c ( ( c − 1 ) ( a − 1 ) ( b − 1 ) a b c ) = x ( a − 1 ) ( b − 1 ) ( c − 1 ) \dfrac{x}{\left(1+\dfrac{a}{b} \right) \left(1+\dfrac{b}{c} \right) \left(1+\dfrac{c}{a} \right)}\\ =\dfrac{x}{a\left(\dfrac{1}{a}+\dfrac{1}{b} \right) b\left(\dfrac{1}{b}+\dfrac{1}{c} \right) c\left(\dfrac{1}{c}+\dfrac{1}{a} \right)}\\ =\dfrac{x}{abc\left(1-\dfrac{1}{c} \right) \left(1-\dfrac{1}{a} \right) \left(1-\dfrac{1}{b} \right)}\\ =\dfrac{x}{abc\left(\dfrac{c-1}{c} \right) \left(\dfrac{a-1}{a} \right) \left(\dfrac{b-1}{b} \right)}\\ =\dfrac{x}{abc\left(\dfrac{(c-1)(a-1)(b-1)}{abc} \right)}\\ =\dfrac{x}{(a-1)(b-1)(c-1)}

We equate the LHS and RHS again:

1 a ( b − 1 ) ( c − 1 ) + 1 b ( a − 1 ) ( c − 1 ) + 1 c ( a − 1 ) ( b − 1 ) = x ( a − 1 ) ( b − 1 ) ( c − 1 ) ( a − 1 ) ( b − 1 ) ( c − 1 ) ( 1 a ( b − 1 ) ( c − 1 ) + 1 b ( a − 1 ) ( c − 1 ) + 1 c ( a − 1 ) ( b − 1 ) ) = x x = ( a − 1 ) ( b − 1 ) ( c − 1 ) a ( b − 1 ) ( c − 1 ) + ( a − 1 ) ( b − 1 ) ( c − 1 ) b ( a − 1 ) ( c − 1 ) + ( a − 1 ) ( b − 1 ) ( c − 1 ) c ( a − 1 ) ( b − 1 ) = a − 1 a + b − 1 b + c − 1 c = 1 − 1 a + 1 − 1 b + 1 − 1 c = 3 − ( 1 a + 1 b + 1 c ) = 3 − 1 x = 2 \dfrac{1}{a(b-1)(c-1)} + \dfrac{1}{b(a-1)(c-1)} + \dfrac{1}{c(a-1)(b-1)} = \dfrac{x}{(a-1)(b-1)(c-1)}\\ (a-1)(b-1)(c-1)\left(\dfrac{1}{a(b-1)(c-1)} + \dfrac{1}{b(a-1)(c-1)} + \dfrac{1}{c(a-1)(b-1)}\right) = x\\ x=\dfrac{(a-1)(b-1)(c-1)}{a(b-1)(c-1)} + \dfrac{(a-1)(b-1)(c-1)}{b(a-1)(c-1)} + \dfrac{(a-1)(b-1)(c-1)}{c(a-1)(b-1)}\\ =\dfrac{a-1}{a} + \dfrac{b-1}{b} + \dfrac{c-1}{c}\\ =1-\dfrac{1}{a} + 1 - \dfrac{1}{b} + 1 - \dfrac{1}{c}\\ =3 - \left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right)\\ =3-1\\ x=\boxed{2}

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Wonderful. I think the question is harder than level 2 algebra

حافظ القران الكريم - 5 years, 1 month ago

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I think so too. It took me more than half an hour to factorize it completely

Hung Woei Neoh - 5 years, 1 month ago

Hint* ,, Let a=b=c

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