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Let $a,b,c$ are positive real numbers and $a^3+b^3+c^3=a^4+b^4+c^4$. Prove that: $$\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+c^3+a^3}+\frac{c}{c^2+a^3+b^3}\ge1$$

My solve: So we have $$\frac{a^2}{a^3+ab^3+ac^3}+\frac{b^2}{b^3+bc^3+ba^3}+\frac{c^2}{c^3+ca^3+cb^3}\ge\frac{(a+b+c)^2}{a^3+b^3+c^3+ab^3+ac^3+bc^3+ba^3+ca^3+cb^3}$$ Now because $a^3+b^3+c^3=a^4+b^4+c^4$ so $$a^3+b^3+c^3+ab^3+ac^3+bc^3+ba^3+ca^3+cb^3$$$$=a^4+b^4+c^4+ab^3+ac^3+bc^3+ba^3+ca^3+cb^3$$$$=(a+b+c)(a^3+b^3+c^3)$$ But I can't prove $a+b+c=a^3+b^3+c^3$. If you have an idea, then please let me know, thank you!

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Hint: For positive $x$, $f(x)=(x-x^3)-2(x^3-x^4)=x(x-1)^2(2x+1)\geqslant 0$. Hence $f(a)+f(b)+f(c) = (a+b+c)-(a^3+b^3+c^3) \geqslant 0$.

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  • $\begingroup$ Thank you, I can do it now $\endgroup$
    – tompi2394
    Nov 5, 2021 at 14:16

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