Distinct real numbers \( a, b, c \) satisfy $ \frac{a - b}{a^3b^3} + \frac{b - c}{b^3c^3} + \frac{c - a}{c^3a^3} = 0. $ Prove that$$ \frac{a^4 + b^4 + c^4}{a^2b^2 + b^2c^2 + c^2a^2}=2$$

@sqing is it ok?

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Rohit-2006 wrote: @sqing is it ok? No:

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Oh sorry I was thinking about something else

Now it's ok sqing?

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Rohit-2006 wrote: Now it's ok sqing? \( a, b, c \) be real numbers.

sqing wrote: Distinct real numbers \( a, b, c \) satisfy $ \frac{a - b}{a^3b^3} + \frac{b - c}{b^3c^3} + \frac{c - a}{c^3a^3} = 0. $ Prove that$$ \frac{a^4 + b^4 + c^4}{a^2b^2 + b^2c^2 + c^2a^2}\textcolor{red}{=2}$$

sqing wrote: Rohit-2006 wrote: Now it's ok sqing? \( a, b, c \) be real numbers. So what's wrong??

Rohit-2006 wrote: sqing wrote: Rohit-2006 wrote: Now it's ok sqing? \( a, b, c \) be real numbers. So what's wrong?? The problem is the differences under the square root can be negative. Using inequalities is not the way to go, I suggest to look what happens with the condition when two of variables are equal.

https://latex.artofproblemsolving.com/miscpdf/lrhbhxkn.pdf?t=1737390722826

$c^3(a-b)+a^3(b-c)+b^3(c-a)=0$ which is $(a-b)(b-c)(c-a)(a+b+c)=0$ So $a+b+c=0$ $a^2+b^2+c^2=-2(ab+bc+ca)$ $a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)=4(a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c))=4(a^2b^2+b^2c^2+c^2a^2)$ $a^4+b^4+c^4=2(a^2b^2+b^2c^2+c^2a^2)$

Sorry.Thank no_room_for_error.

Nice.

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RagvaloD wrote: $c^3(a-b)+a^3(b-c)+b^3(c-a)=0$ which is $(a-b)(b-c)(c-a)(a+b+c)=0$ So $a+b+c=0$ $a^2+b^2+c^2=-2(ab+bc+ca)$ $a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)=4(a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c))=4(a^2b^2+b^2c^2+c^2a^2)$ $a^4+b^4+c^4=2(a^2b^2+b^2c^2+c^2a^2)$ Indeed ,this it is more nice solution!