Proving Equality in Real Numbers: The Case of a+b+c=abc

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SUMMARY

The discussion centers on proving the equality for real numbers \(a\), \(b\), and \(c\) under the condition \(a+b+c=abc\). The derived equation is \(\frac{a^2}{1-a^2}+\frac{b^2}{1-b^2}+\frac{c^2}{1-c^2}=\frac{4abc}{(1-a^2)(1-b^2)(1-c^2)}\). The proof utilizes trigonometric identities, specifically relating \(a\), \(b\), and \(c\) to the tangent function, leading to the conclusion that the sum of the transformed variables equals the product of the original variables scaled by a factor of four. The approach taken by participants aligns, confirming the validity of the method used.

PREREQUISITES
  • Understanding of trigonometric identities, particularly tangent functions.
  • Familiarity with algebraic manipulation involving real numbers.
  • Knowledge of the properties of equality and inequalities in mathematics.
  • Basic understanding of mathematical proofs and logical reasoning.
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  • Study the properties of tangent functions and their applications in proofs.
  • Explore advanced algebraic techniques for manipulating equations involving multiple variables.
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anemone
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Let $a, b$ and $c$ be real numbers, all different from -1 and 1, such that $a+b+c=abc$. Prove that $$\frac{a^2}{1-a^2}+\frac{b^2}{1-b^2}+\frac{c^2}{1-c^2}=\frac{4abc}{(1-a^2)(1-b^2)(1-c^2)}$$
 
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anemone said:
Let $a, b$ and $c$ be real numbers, all different from -1 and 1, such that $a+b+c=abc$. Prove that $$\frac{a^2}{1-a^2}+\frac{b^2}{1-b^2}+\frac{c^2}{1-c^2}=\frac{4abc}{(1-a^2)(1-b^2)(1-c^2)}$$

(a + b+ c) = abc

so a = - (b+ c)/(1-bc)

if a = tan A , b = tan B , c = tan C

then tan A = - tan (B+C)

or A + B + C = npi

so 2A + 2B + 2C= 2npi

so tan 2A = tan (2npi- 2B - 2C)

or tan 2A = - tan (2B + 2C)
= (tan 2B + tan 2C)/( tan 2B tan 2C -1)

or tan 2A tan 2B tan 2C - tan 2A = tan 2B + tan 2C
or tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C

or
2tan A/(1-tan ^2A) + 2 tan B/(1-tan ^2 B) + 2 tanC /(1 - tan^2C) = 8 tan A tanC tanC/(( 1-tan ^2A)(1-tan ^2 B)(1- tan ^2C)

or
tan A/(1-tan ^2A) + tan B/(1-tan ^2 B) + tanC /(1 - tan^2C) = 4 tan A tanC tanC/(( 1-tan ^2A)(1-tan ^2 B)(1- tan ^2C)

or
a/(1-a^2) + b/(1-b^2) + c/(1 - c^2) = 4abc/(( 1-a^2)(1-b^2)(1- c^2))
 
kaliprasad said:
(a + b+ c) = abc

so a = - (b+ c)/(1-bc)

if a = tan A , b = tan B , c = tan C

then tan A = - tan (B+C)

or A + B + C = npi

so 2A + 2B + 2C= 2npi

so tan 2A = tan (2npi- 2B - 2C)

or tan 2A = - tan (2B + 2C)
= (tan 2B + tan 2C)/( tan 2B tan 2C -1)

or tan 2A tan 2B tan 2C - tan 2A = tan 2B + tan 2C
or tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C

or
2tan A/(1-tan ^2A) + 2 tan B/(1-tan ^2 B) + 2 tanC /(1 - tan^2C) = 8 tan A tanC tanC/(( 1-tan ^2A)(1-tan ^2 B)(1- tan ^2C)

or
tan A/(1-tan ^2A) + tan B/(1-tan ^2 B) + tanC /(1 - tan^2C) = 4 tan A tanC tanC/(( 1-tan ^2A)(1-tan ^2 B)(1- tan ^2C)

or
a/(1-a^2) + b/(1-b^2) + c/(1 - c^2) = 4abc/(( 1-a^2)(1-b^2)(1- c^2))

Thanks again for participating, kaliprasad and just so you know, you and I used the same approach to solve this problem!:)
 

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