Evaluate (a²+b²+c²)/(ab+bc+ca)

[anemone]

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]

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.

Spoiler

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

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

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

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.

 

[anemone]

Hello.

Spoiler

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

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

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

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! :)