Replace $(b+c)^2$ with $4bc$ $\sum \frac{a}{1+4bc} \leq \frac{3(a^2+b^2+c^2)}{a^2+b^2+c^2+12abc}$ $\Leftrightarrow $ $\sum [ a( \frac{3a}{a^2+b^2+c^2+12abc} - \frac{1}{1+4bc}) ] \geq 0 $ $\Leftrightarrow $ $ \sum [ \frac{3a^2 - a(a^2 +b^2 +c^2)}{1+4bc} ] = \sum [ \frac{a^2 (a+b+c) - a(a^2 +b^2 +c^2)}{1+4bc} ] \geq 0$ $\Leftrightarrow $ $\sum [\frac{ab(a-b)+ac(a-c)}{1+4bc}] \geq 0$ or $\sum [\frac{ab(a-b)}{1+4bc} + \frac{ab(b-a)}{1+4ac}] \geq 0 $ Which can be written $\sum [\frac{4abc(a-b)^2}{(1+4bc)(1+4ac)}] \geq 0 $

goodar2006 wrote: for positive real numbers $a,b$ and $c$ we have $a+b+c=3$. prove $\frac{a}{1+(b+c)^2}+\frac{b}{1+(a+c)^2}+\frac{c}{1+(a+b)^2}\le \frac{3(a^2+b^2+c^2)}{a^2+b^2+c^2+12abc}$. Write the inequality as \[{\left[ a-\frac{a}{1+(b+c)^2}\right] +\left[b-\frac{b}{1+(c+a)^2}\right]+\left[ c-\frac{c}{1+(a+b)^2}\right] \ge 3-\frac{3(a^2+b^2+c^2)}{a^2+b^2+c^2+12abc},}\] \[\frac{a(b+c)^2}{1+(b+c)^2}+\frac{b(c+a)^2}{1+(c+a)^2}+\frac{c(a+b)^2}{1+(a+b)^2} \ge \frac{36abc}{a^2+b^2+c^2+12abc}.\] Using the Cauchy-Schwarz inequality, we get \[\left[ \sum \frac{a(b+c)^2}{1+(b+c)^2}\right] \left\{ \sum \frac{a\left[1+(b+c)^2\right]}{(b+c)^2}\right\} \ge \left(\sum a\right)^2 =9.\] Therefore, it suffices to prove that \[\frac{9}{\frac{a\left[1+(b+c)^2\right]}{(b+c)^2}+\frac{b\left[1+(c+a)^2\right]}{(c+a)^2}+\frac{c\left[1+(a+b)^2\right]}{(a+b)^2}} \ge \frac{36abc}{a^2+b^2+c^2+12abc},\] or \[\frac{a^2+b^2+c^2+12abc}{abc} \ge 4\left\{\frac{a\left[1+(b+c)^2\right]}{(b+c)^2}+\frac{b\left[1+(c+a)^2\right]}{(c+a)^2}+\frac{c\left[1+(a+b)^2\right]}{(a+b)^2}\right\}.\] Since \[\begin{aligned} \frac{a\left[1+(b+c)^2\right]}{(b+c)^2}+\frac{b\left[1+(c+a)^2\right]}{(c+a)^2}+\frac{c\left[1+(a+b)^2\right]}{(a+b)^2}&=\left[ \frac{a}{(b+c)^2}+\frac{b}{(c+a)^2}+\frac{c}{(a+b)^2}\right]+(a+b+c) \\ &= \frac{a}{(b+c)^2}+\frac{b}{(c+a)^2}+\frac{c}{(a+b)^2}+3, \end{aligned} \] the above inequality can be written as \[\frac{a^2+b^2+c^2}{abc} \ge 4\left[ \frac{a}{(b+c)^2}+\frac{b}{(c+a)^2}+\frac{c}{(a+b)^2}\right] ,\] which is true because \[4\left[ \frac{a}{(b+c)^2}+\frac{b}{(c+a)^2}+\frac{c}{(a+b)^2}\right] \le 4\left(\frac{a}{4bc}+\frac{b}{4ca}+\frac{c}{4ab}\right) =\frac{a^2+b^2+c^2}{abc}.\] The proof is completed. $\blacksquare$

Who are you man !? May I borrow your mind for next exam ? Really nice solution Mr.Can Hang .

really wonderful!!!! I can't say anything

Indeed You have a good command of Cauchy-Schwarz Inequality.

can_hang2007 wrote: goodar2006 wrote: for positive real numbers $a,b$ and $c$ we have $a+b+c=3$. prove $\frac{a}{1+(b+c)^2}+\frac{b}{1+(a+c)^2}+\frac{c}{1+(a+b)^2}\le \frac{3(a^2+b^2+c^2)}{a^2+b^2+c^2+12abc}$. Write the inequality as \[{\left[ a-\frac{a}{1+(b+c)^2}\right] +\left[b-\frac{b}{1+(c+a)^2}\right]+\left[ c-\frac{c}{1+(a+b)^2}\right] \ge 3-\frac{3(a^2+b^2+c^2)}{a^2+b^2+c^2+12abc},}\] \[\frac{a(b+c)^2}{1+(b+c)^2}+\frac{b(c+a)^2}{1+(c+a)^2}+\frac{c(a+b)^2}{1+(a+b)^2} \ge \frac{36abc}{a^2+b^2+c^2+12abc}.\] Using the Cauchy-Schwarz inequality, we get \[\left[ \sum \frac{a(b+c)^2}{1+(b+c)^2}\right] \left\{ \sum \frac{a\left[1+(b+c)^2\right]}{(b+c)^2}\right\} \ge \left(\sum a\right)^2 =9.\] Therefore, it suffices to prove that \[\frac{9}{\frac{a\left[1+(b+c)^2\right]}{(b+c)^2}+\frac{b\left[1+(c+a)^2\right]}{(c+a)^2}+\frac{c\left[1+(a+b)^2\right]}{(a+b)^2}} \ge \frac{36abc}{a^2+b^2+c^2+12abc},\] or \[\frac{a^2+b^2+c^2+12abc}{abc} \ge 4\left\{\frac{a\left[1+(b+c)^2\right]}{(b+c)^2}+\frac{b\left[1+(c+a)^2\right]}{(c+a)^2}+\frac{c\left[1+(a+b)^2\right]}{(a+b)^2}\right\}.\] Since \[\begin{aligned} \frac{a\left[1+(b+c)^2\right]}{(b+c)^2}+\frac{b\left[1+(c+a)^2\right]}{(c+a)^2}+\frac{c\left[1+(a+b)^2\right]}{(a+b)^2}&=\left[ \frac{a}{(b+c)^2}+\frac{b}{(c+a)^2}+\frac{c}{(a+b)^2}\right]+(a+b+c) \\ &= \frac{a}{(b+c)^2}+\frac{b}{(c+a)^2}+\frac{c}{(a+b)^2}+3, \end{aligned} \] the above inequality can be written as \[\frac{a^2+b^2+c^2}{abc} \ge 4\left[ \frac{a}{(b+c)^2}+\frac{b}{(c+a)^2}+\frac{c}{(a+b)^2}\right] ,\] which is true because \[4\left[ \frac{a}{(b+c)^2}+\frac{b}{(c+a)^2}+\frac{c}{(a+b)^2}\right] \le 4\left(\frac{a}{4bc}+\frac{b}{4ca}+\frac{c}{4ab}\right) =\frac{a^2+b^2+c^2}{abc}.\] The proof is completed. $\blacksquare$ Infact , $f(x)=\frac{x}{1+x}$ is increasing when $x>0$ , so we have \[\frac{a(b+c)^2}{1+(b+c)^2}+\frac{b(c+a)^2}{1+(c+a)^2}+\frac{c(a+b)^2}{1+(a+b)^2}\ge\frac{4abc}{1+4bc}+\frac{4abc}{1+4ca}+\frac{4abc}{1+4ab}\]We just need to prove \[\frac{4abc}{1+4bc}+\frac{4abc}{1+4ca}+\frac{4abc}{1+4ab}\ge\frac{36abc}{a^2+b^2+c^2+12abc}=\frac{36abc}{\sum_{\mathrm{cyc}}a^2(1+4bc)}\]which is obviously by Cauchy-Schwarz \[\sum_{\mathrm{cyc}}\frac{1}{1+4bc}\sum_{\mathrm{cyc}}a^2(1+4bc)\ge (a+b+c)^2=9\]

we can use ABC theorem to solve this problem \[\sum\frac{a(b+c)^2}{1+(b+c)^2}\ge\frac{36abc}{a^2+b^2+c^2+12abc}.\] By Cauchy inequality \[\sum\frac{a(b+c)^2}{1+(b+c)^2}\ge\frac{4(ab+bc+ca)^2}{a+b+c+a^{2}b+ab^2+b^{2}c+bc^2+c^{2}a+ca^2+6abc}.\] Enough to show \[\frac{4(ab+bc+ca)^2}{a+b+c+a^{2}b+ab^2+b^{2}c+bc^2+c^{2}a+ca^2+6abc}\ge\frac{36abc}{a^2+b^2+c^2+12abc}.\] let ab+bc+ca=q, abc=r and we expand above it we can get \[27r^2+27qr+27r-12rq^2-9q^2+2q^3\le\ 0 .\] it is about r function and convex form so we can use ABC theorem so we can solve this inequality about c=0 or a=b=x(but 0<x<3/2) and this is easy thing Also this is my first latex work ㅡㅡ

Sorry for being an idiot , but what is the $ABC$ theorem ? Quote: By Cauchy inequality \[ \sum\frac{a(b+c)^{2}}{1+(b+c)^{2}}\ge\frac{4(ab+bc+ca)^{2}}{a+b+c+a^{2}b+ab^{2}+b^{2}c+bc^{2}+c^{2}a+ca^{2}}. \] How ? I think you miss $6abc$ in denominator ...

Oh I missed 6abc when I typed the eqution.. sorry ABC theorem is when we suppose p=a+b+c, q=ab+bc+ca, r=abc (a,b,c is positive real number) we want to show \[f(a,b,c)\le\ 0 \] when we represent f(a,b,c) by p,q,r then if this function is convex function about r we only check \[f(a,a,b), f(a,b,0) \le\ 0 \]