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?

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).
