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For $a,b,c \in (0,\infty).$ Prove$:$ $$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ac}+\frac{c(b+a)}{c^2+ba}\geqq 1+\frac{16abc}{(a+b)(b+c)(c+a)} $$

My proof by SOS$:$

$$ \left( {a}^{2}+bc \right) \left( ac+{b}^{2} \right) \left( ab+{c}^{ 2} \right) \left( a+b \right) \left( b+c \right) \left( c+a \right)\, \cdot \,(\text{LHS}-\text{RHS})$$

$$=\frac{5}{4} abc \sum\limits_{cyc} c^2 (a+b-2c)^2 (a-b)^2 +\frac{1}{4} \sum\limits_{cyc} {c}^{3} \left( 4\,{a}^{2}+3\,ab+4\,{b}^{2} \right) \left( a-b \right) ^{4}$$

However$,$ it's hard to find this SOS's form without computer.

So I am looking for alternative solution without $uvw.$ Thanks very much!

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4 Answers 4

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We have $$\frac{a(b+c)}{a^2+bc}+\frac{b(a+c)}{b^2+ca}+\frac{c(a+b)}{c^2+ab} - 1 - \frac{16abc}{(a+b)(b+c)(a+c)}$$ $$=\sum{\frac{(c-a)^2(c-b)^2}{(c^2+ab)(b+c)(c+a)}}\geqslant 0.$$

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  • $\begingroup$ You found this by computer, right? $\endgroup$
    – NKellira
    Jun 6, 2020 at 13:04
  • $\begingroup$ (+1) Simpler SOS. $\endgroup$
    – River Li
    Jun 6, 2020 at 13:48
  • $\begingroup$ @RiverLi Did you found any SOS for it$?$ $\endgroup$
    – NKellira
    Jun 6, 2020 at 13:57
  • $\begingroup$ @knvy144444 I cannot find so simple form. $\endgroup$
    – River Li
    Jun 6, 2020 at 14:02
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We can prove this inequality as an inequality in the Schur's form typing.

Indeed, we need to prove that: $$\sum_{cyc}\frac{a(b+c)}{a^2+bc}\geq1+\frac{16abc}{\prod\limits_{cyc}(a+b)}$$ or $$2-\frac{16abc}{\prod\limits_{cyc}(a+b)}\geq\sum_{cyc}\left(1-\frac{a(b+c)}{a^2+bc}\right)$$ or $$\frac{2\sum\limits_{cyc}(a^2b+a^2c-2abc)}{\prod\limits_{cyc}(a+b)}\geq\sum_{cyc}\frac{(a-b)(a-c)}{a^2+bc}$$ or $$\frac{2\sum\limits_{cyc}(b+c)(a-b)(a-c)}{\prod\limits_{cyc}(a+b)}\geq\sum_{cyc}\frac{(a-b)(a-c)}{a^2+bc}$$ or $$\sum_{cyc}(a-b)(a-c)\left(\frac{2}{(a+b)(a+c)}-\frac{1}{a^2+bc}\right)\geq0$$ or $$\sum_{cyc}\frac{(a-b)^2(a-c)^2}{(a+b)(a+c)(a^2+bc)}\geq0$$ and we are done!

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    $\begingroup$ It is very nice. $\endgroup$
    – River Li
    Jun 7, 2020 at 2:13
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$$LHS-RHS=\frac{4abc(b-c)^{2}(c-a)^{2}(a-b)^{2}}{(b+c)(c+a)(a+b)(a^{2}+bc)(b^{2}+ca)(c^{2}+ab)}+\frac{(a^{2}+b^{2}+c^{2}-bc-ca-ab)(b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}-a^{2}bc-b^{2}ca-c^{2}ab)}{(a^{2}+bc)(b^{2}+ca)(c^{2}+ab)}\geq 0.$$

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  • $\begingroup$ (+1) Very nice SOS. $\endgroup$
    – River Li
    Oct 10, 2021 at 1:38
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Here is an alternative proof using AM-GM.

Split into two cases:

  • $abc=0:$ Let $a=0; b=c>0$ the OP is obviously true $$\frac{b}{c}+\frac{c}{b}\ge 1.$$
  • $abc>0.$ By AM-GM \begin{align*} \dfrac{a(b+c)}{a^2+bc}+1&\ge \dfrac{2a(b+c)}{\sqrt{(a^2+bc)(ab+ac)}}\\ &\ge \dfrac{4a(b+c)}{a^2+bc+ab+ca}\\ &=\dfrac{4a(b+c)}{(a+b)(c+a)}. \end{align*} Similarly, we obtain \begin{align*} \sum\limits_{cyc}\dfrac{a(b+c)}{a^2+bc}+3 &\ge \dfrac{4a(b+c)^2+4b(c+a)^2+4c(a+b)^2}{(a+b)(b+c)(c+a)}\\ &=4+\dfrac{16abc}{(a+b)(b+c)(c+a)}. \end{align*} The desired result follows. We end proof here. Equality holds at $a=b=c>0.$
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