inequality

Let $a, b, c$ be positive real numbers satisfying $a^3+b^3+c^3=a^4+b^4+c^4$. Show that
\[ \frac{a}{a^2+b^3+c^3}+\frac{b}{a^3+b^2+c^3}+\frac{c}{a^3+b^3+c^2} \geq 1 \]

Solution :$\sum \frac{a}{a^2+b^3+c^3}=\sum \frac{a^2}{a^3+ab^3+ac^3}$$\geq \frac{(a+b+c)^2}{a^3+b^3+c^3+\sum_{sym}a^3b}$$= \frac{(a+b+c)^2}{a^4+b^4+c^4+\sum_{sym}a^3b}$$= \frac{(a+b+c)^2}{(a+b+c)(a^3+b^3+c^3)}$$= \frac{(a+b+c)(a^4+b^4+c^4)^2}{(a^3+b^3+c^3)^3}$

But the last expression is obviously $\geq 1$ by Holder inequality . so , we are done

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