MHB Evaluate (a²+b²+c²)/(ab+bc+ca)

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The discussion focuses on evaluating the expression (a²+b²+c²)/(ab+bc+ca) under specific conditions involving real numbers a, b, and c. It states that a²/b + b²/c + c²/a equals a²/c + b²/a + c²/b, while the sum a/b + b/c + c/a does not equal a/c + b/a + c/b. Participants engage in exploring the implications of these equations to derive the value of the expression. The conversation highlights the mathematical relationships and potential simplifications that arise from the given conditions. The conclusion emphasizes the importance of these relationships in determining the final value.
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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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