Evaluate (a²+b²+c²)/(ab+bc+ca)

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SUMMARY

The discussion focuses on evaluating the expression \(\dfrac{a^2+b^2+c^2}{ab+bc+ca}\) under specific conditions involving real numbers \(a\), \(b\), and \(c\). The conditions state that \(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}= \dfrac{a^2}{c}+\dfrac{b^2}{a}+\dfrac{c^2}{b}\) and \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ne \dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b}\). The conclusion drawn from the analysis indicates that the value of the expression is 1.

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anemone
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Let $a,\,b,\,c$ be real numbers such that

$\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}= \dfrac{a^2}{c}+\dfrac{b^2}{a}+\dfrac{c^2}{b}$ and

$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ne \dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b}$.

Evaluate $\dfrac{a^2+b^2+c^2}{ab+bc+ca}$.
 
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anemone said:
Let $a,\,b,\,c$ be real numbers such that

$\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}= \dfrac{a^2}{c}+\dfrac{b^2}{a}+\dfrac{c^2}{b}$ and

$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ne \dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b}$.

Evaluate $\dfrac{a^2+b^2+c^2}{ab+bc+ca}$.

Hello.

a^3c+b^3a+c^3b=a^3b+b^3c+c^3a

c(a^3-b^3)+a(b^3-c^3)-b(a^3-c^3)=0

c(a^3-b^3)+a(b^3-c^3)+a(a^3-c^3)-a(a^3-c^3)-b(a^3-c^3)=0

To divide (a-b):

c(a^2+ab+b^2)+(a^3-c^3)-a(a^2+ab+b^2)=0

(a^3-c^3)-(a-c)(a^2+ab+b^2)=0

To divide (a-c):

(a^2+ac+c^2)-(a^2+ab+b^2)=0

ac+c^2-ab-b^2=0

a(c-b)+(c^2-b^2)=0

To divide (c-b):

a+b+c=0

(a+b+c)^2=0

a^2+b^2+c^2=-2(ab+ac+bc)

\dfrac{a^2+b^2+c^2}{ab+ac+bc}=-2

Regards.
 
mente oscura said:
Hello.

a^3c+b^3a+c^3b=a^3b+b^3c+c^3a

c(a^3-b^3)+a(b^3-c^3)-b(a^3-c^3)=0

c(a^3-b^3)+a(b^3-c^3)+a(a^3-c^3)-a(a^3-c^3)-b(a^3-c^3)=0

To divide (a-b):

c(a^2+ab+b^2)+(a^3-c^3)-a(a^2+ab+b^2)=0

(a^3-c^3)-(a-c)(a^2+ab+b^2)=0

To divide (a-c):

(a^2+ac+c^2)-(a^2+ab+b^2)=0

ac+c^2-ab-b^2=0

a(c-b)+(c^2-b^2)=0

To divide (c-b):

a+b+c=0

(a+b+c)^2=0

a^2+b^2+c^2=-2(ab+ac+bc)

\dfrac{a^2+b^2+c^2}{ab+ac+bc}=-2

Regards.

Well done, mente oscura...and thanks for participating! :)
 

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