Prove the inequality $\sum_{cyc}\frac{a}{1+\left(b+c\right)^2}\le \frac{3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2+12abc}$

Question

Let $a>0$, $b>0$ and $c>0$ such that $a+b+c=3$. Prove the inequality $$\frac{a}{1+\left(b+c\right)^2}+\frac{b}{1+\left(c+a\right)^2}+\frac{c}{1+\left(a+b\right)^2}\le \frac{3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2+12abc}.$$

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$$LHS=\frac{a}{1+\left(3-a\right)^2}+\frac{b}{1+\left(3-b\right)^2}+\frac{c}{1+\left(3-c\right)^2}$$

We have inequality: $$\frac{a}{1+\left(3-a\right)^2}\le \frac{9}{25}a-\frac{4}{25}\Leftrightarrow -\frac{\left(a-1\right)^2\left(9a-40\right)}{25\left(\left(a-3\right)^2+1\right)}\le 0\forall 0<a\le 1$$

$$\Rightarrow LHS\le \frac{9}{25}\left(a+b+c\right)-\frac{4}{25}\cdot 3=\frac{3}{5}$$

Need prove: $$\frac{3}{5}\le \frac{3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2+12abc}$$

$$\Leftrightarrow a^2+b^2+c^2+12abc\le 5\left(a^2+b^2+c^2\right)$$

$$\Leftrightarrow 12abc\le 4\left(a^2+b^2+c^2\right)\Leftrightarrow 3abc\le a^2+b^2+c^2$$

By AM-GM: $$3abc\le \frac{\left(a+b+c\right)^3}{9}=3=\frac{\left(a+b+c\right)^2}{3}\le a^2+b^2+c^2$$

Right or Wrong. I think it's wrong. Help me

Answer 1

We need to prove that $$\sum_{cyc}\left(\frac{a}{1+(b+c)^2}-a\right)\leq\frac{3(a^2+b^2+c^2)}{a^2+b^2+c^2+12abc}-3$$ or $$\sum_{cyc}\frac{a(b+c)^2}{1+(b+c)^2}\geq\frac{36abc}{a^2+b^2+c^2+12abc}.$$ Now, by C-S $$\sum_{cyc}\frac{a(b+c)^2}{1+(b+c)^2}=\sum_{cyc}\frac{a}{1+\frac{1}{(b+c)^2}}=$$ $$=\sum_{cyc}\frac{a^2}{a+\frac{a}{(b+c)^2}}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}\left(a+\frac{a}{(b+c)^2}\right)}=\frac{9}{3+\sum\limits_{cyc}\frac{a}{(b+c)^2}}.$$ Thus, it remains to prove that $$a^2+b^2+c^2+12abc\geq4abc\left(3+\sum\limits_{cyc}\frac{a}{(b+c)^2}\right)$$ or $$a^2+b^2+c^2\geq4abc\sum\limits_{cyc}\frac{a}{(b+c)^2}$$ or $$\sum_{cyc}\left(\frac{a}{bc}-\frac{4a}{(b+c)^2}\right)\geq0$$ or $$\sum_{cyc}\frac{a(b-c)^2}{bc(b+c)^2}\geq0.$$ Done!

Answer 2

We need to prove that $$\sum_{cyc}\left(\frac{a}{1+(b+c)^2}-a\right)\leq\frac{3(a^2+b^2+c^2)}{a^2+b^2+c^2+12abc}-3$$ or $$\sum_{cyc}\frac{a(b+c)^2}{1+(b+c)^2}\geq\frac{36abc}{a^2+b^2+c^2+12abc},$$ which seems better, but I did not find something nice here.

By the way, our inequality is obviously true after homogenization and full expanding.

Indeed, we need to prove that $$\sum_{cyc}\frac{a}{(a+b+c)^2+9(b+c)^2}\leq\frac{a^2+b^2+c^2}{(a^2+b^2+c^2)(a+b+c)+36abc}$$ or $$\sum_{sym}(10a^7b+130a^6b^2+250a^5b^3+130a^4b^4-68a^6bc-159a^5b^2c+255a^4b^3c-225a^4b^2c^2-323a^3b^3c^2)\geq0,$$ which is true by Muirhead: $$\sum_{sym}10a^7b\geq\sum_{sym}10a^6b^2;$$ $$\sum_{sym}68a^6b^2\geq\sum_{sym}68a^6bc;$$ $$\sum_{sym}72a^6b^2\geq\sum_{sym}72a^5b^3;$$ $$\sum_{sym}159a^5b^3\geq\sum_{sym}159a^5b^2c;$$ $$\sum_{sym}163a^5b^3\geq\sum_{sym}163a^4b^2c^2;$$ $$\sum_{sym}62a^4b^4\geq\sum_{sym}62a^4b^2c^2;$$ $$\sum_{sym}68a^4b^4\geq\sum_{sym}68a^3b^3c^2$$ and $$\sum_{sym}255a^4b^3c\geq\sum_{sym}255a^3b^3c^2.$$ After summing of these inequalities we'll get the needed inequality.

Done!

Answer 3

The following quantity is clearly positive \begin{eqnarray*} 90 \sum_{perms} a^5b(a-c)^2+1170 \sum_{cyclic} a^6(b-c)^2+2250 \sum_{perms} a^5 b(b-c)^2+1476 \sum_{cyclic} abc^2(ab-c^2)^2+729abc \sum_{perms} a^4(b-c)^4+6246 \sum_{cyclic} a^4 bc (b-c)^2 +2187 a^2 b^2 c^2 \sum_{cyclic} (a-b)^2 \geq 0 . \end{eqnarray*} This can be rearranged to \begin{eqnarray*} \sum_{perms} a((a+b+c)^2+9(c+a)^2)((a+b+c)^2+9(a+b)^2)((a+b+c)(a^2+b^2+c^2)+36abc) \leq 3(a^2+b^2+c^2)((a+b+c)^2+9(c+a)^2)((a+b+c)^2+9(a+b)^2)((a+b+c)^2+9(b+c)^2) \end{eqnarray*} Now substitute $a+b+c=3$ and divide by $(1+(c+a)^2)(1+(a+b)^2)(1+(b+c)^2)((a^2+b^2+c^2)+12abc)$ and we have \begin{eqnarray*} \frac{ a}{1+(b+c)^2}+\frac{ b}{1+(c+a)^2}+\frac{ c}{1+(a+b)^2} \leq \frac{3(a^2+b^2+c^2)}{a^2+b^2+c^2+12abc} \end{eqnarray*}

The above algebra can be justified by running the following reduce program

( (a^2+b^2+c^2)*((a+b+c)^2+9*(c+a)^2)*((a+b+c)^2+9*(a+b)^2)*((a+b+c)^2+9*(b+c)^2)-
 a*((a+b+c)^2+9*(c+a)^2)*((a+b+c)^2+9*(a+b)^2)*((a+b+c)*(a^2+b^2+c^2)+36*a*b*c)-
  b*((a+b+c)^2+9*(c+b)^2)*((a+b+c)^2+9*(a+b)^2)*((a+b+c)*(a^2+b^2+c^2)+36*a*b*c)-
 c*((a+b+c)^2+9*(c+a)^2)*((a+b+c)^2+9*(c+b)^2)*((a+b+c)*(a^2+b^2+c^2)+36*a*b*c))-(
 90*(a^5*b*(a-c)^2+b^5*c*(b-a)^2+c^5*a*(c-b)^2+a^5*c*(a-b)^2+b^5*a*(b-c)^2+c^5*b*(c-a)^2)+
 1170*(a^6*(b-c)^2+b^6*(c-a)^2+c^6*(a-b)^2)+
 2250*(a^5*(b+c)*(b-c)^2+b^5*(c+a)*(c-a)^2+c^5*(a+b)*(a-b)^2)+
 1476*(a*b*c^2*(a*b-c^2)^2+b*c*a^2*(b*c-a^2)^2+c*a*b^2*(c*a-b^2)^2)+
 729*(a*b*c*(b+c)*(a^2-b*c)^2+b*c*a*(c+a)*(b^2-a*c)^2+c*a*b*(a+b)*(c^2-b*a)^2)+
 0*(a^2*b^2*(a*b-c^2)^2+b^2*c^2*(b*c-a^2)^2+c^2*a^2*(c*a-b^2)^2)+
 1170*(a^4*(b-c)^4+b^4*(c-a)^4+c^4*(a-b)^4)+
 6246*(a^4*b*c*(b-c)^2+b^4*c*a*(c-a)^2+c^4*a*b*(a-b)^2)+
 2187*a^2*b^2*c^2*((a-b)^2+(b-c)^2+(c-a)^2)
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