<s>[INST] Prove that $\sum\limits_{cyc}{} \frac{a^3}{b+c} \geq \sum_\limits{cyc}{} \frac{a^4+b^3c-b^2c^2+bc^3}{(a+b)(a+c)}$ If $ a,b,c>0 $, prove that :$$\sum_{cyc}{} \frac{a^3}{b+c} \geq \sum_{cyc}{} \frac{a^4+b^3c-b^2c^2+bc^3}{(a+b)(a+c)}.$$
my attempt:
After uniting the denominator and dividing by$(a+b)(a+c)(b+c)$
$\Leftrightarrow$$a^5+a^4b+a^4c+a^3bc+b^5+b^4a+b^4c+b^3ac+c^5+c^4b+c^4a+c^3ab\geq a^4b+a^4c+b^4c+b^3c^2-b^3c^2-b^2c^3+b^2c^3+bc^4+b^4a+b^4c+a^2c^3+ac^4-a^3c^2-a^2c^3+a^4c+a^3c^2+c^4a+c^4b+a^4b+a^3b^2-a^3b^2-a^2b^2+ab^4+a^2b^3$$\Leftrightarrow$$a^4(a-c)+b^4(b-c)+c^4(c-a)+a^3(bc-ab)+b^3(ac-ab)+c^3(ab-cb)\geq 0$$\Leftrightarrow$$a^3(a-c)(a-b)+b^3(b-c)(b-a)+c^3(c-a)(c-b)\geq 0$
and this last expression is true(it is  schur's inequality).
does my attempt is true?
 [/INST] Your proof is OK. But it is simpler to manipulate cyclic sum.
It suffices to prove that
$$\sum_{\mathrm{cyc}} [a^3(a + b)(a + c) - (a^4 + b^3c - b^2c^2 + bc^3)(b + c)] \ge 0$$
or
$$\sum_{\mathrm{cyc}} (a^5 + a^3bc - b^4c - bc^4) \ge 0$$
or (because $\sum_{\mathrm{cyc}} b^4c = \sum_{\mathrm{cyc}} a^4 b$
and $\sum_{\mathrm{cyc}} bc^4 = \sum_{\mathrm{cyc}} c a^4$)
$$\sum_{\mathrm{cyc}} (a^5 + a^3bc - a^4b - ca^4) \ge 0$$
or
$$\sum_{\mathrm{cyc}} a^3(a - b)(a - c) \ge 0$$
which is true (Schur).
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