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Evaluate (a²+b²+c²)/(ab+bc+ca)

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,756
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 $\dfrac{a^2+b^2+c^2}{ab+bc+ca}$.
 

mente oscura

Well-known member
Nov 29, 2013
172
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.

[tex]a^3c+b^3a+c^3b=a^3b+b^3c+c^3a[/tex]

[tex]c(a^3-b^3)+a(b^3-c^3)-b(a^3-c^3)=0[/tex]

[tex]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[/tex]

To divide (a-b):

[tex]c(a^2+ab+b^2)+(a^3-c^3)-a(a^2+ab+b^2)=0[/tex]

[tex](a^3-c^3)-(a-c)(a^2+ab+b^2)=0[/tex]

To divide (a-c):

[tex](a^2+ac+c^2)-(a^2+ab+b^2)=0[/tex]

[tex]ac+c^2-ab-b^2=0[/tex]

[tex]a(c-b)+(c^2-b^2)=0[/tex]

To divide (c-b):

[tex]a+b+c=0[/tex]

[tex](a+b+c)^2=0[/tex]

[tex]a^2+b^2+c^2=-2(ab+ac+bc)[/tex]

[tex]\dfrac{a^2+b^2+c^2}{ab+ac+bc}=-2[/tex]



Regards.
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,756
Hello.

[tex]a^3c+b^3a+c^3b=a^3b+b^3c+c^3a[/tex]

[tex]c(a^3-b^3)+a(b^3-c^3)-b(a^3-c^3)=0[/tex]

[tex]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[/tex]

To divide (a-b):

[tex]c(a^2+ab+b^2)+(a^3-c^3)-a(a^2+ab+b^2)=0[/tex]

[tex](a^3-c^3)-(a-c)(a^2+ab+b^2)=0[/tex]

To divide (a-c):

[tex](a^2+ac+c^2)-(a^2+ab+b^2)=0[/tex]

[tex]ac+c^2-ab-b^2=0[/tex]

[tex]a(c-b)+(c^2-b^2)=0[/tex]

To divide (c-b):

[tex]a+b+c=0[/tex]

[tex](a+b+c)^2=0[/tex]

[tex]a^2+b^2+c^2=-2(ab+ac+bc)[/tex]

[tex]\dfrac{a^2+b^2+c^2}{ab+ac+bc}=-2[/tex]



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