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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?

River Li
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  • please verifie if it right,and also there is an other problem which is closed without any reason ,it's link: https://math.stackexchange.com/q/4485479/1069990 –  Jul 03 '22 at 22:35
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    Right attempt. Nothing wrong. But, one thing is that yours is a bit hard to read, and also complicated. I recommend you regard your solution and find an easy, simple way. – RDK Jul 04 '22 at 03:18
  • @RDK okey thank you –  Jul 04 '22 at 05:58

1 Answers1

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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).

River Li
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