kunny wrote: Find the maximum value of real number $k$ such that \[\frac{a}{1+9bc+k(b-c)^2}+\frac{b}{1+9ca+k(c-a)^2}+\frac{c}{1+9ab+k(a-b)^2}\geq \frac{1}{2}\] holds for all non-negative real numbers $a,\ b,\ c$ satisfying $a+b+c=1$. Let $a=b=\frac{1}{2}$ and $c=0$. Hence, $k\leq4$ and it remains to prove that $\sum_{cyc}\frac{a}{1+9bc+4(b-c)^2}\geq\frac{1}{2}$, which is SOS.

kunny wrote: What's SOS, arqady^^ It's the following reasoning. Let $a\geq b\geq c$. Hence, $\sum_{cyc}\frac{a}{1+9bc+4(b-c)^2}-\frac{1}{2}=\sum_{cyc}\left(\frac{a}{(a+b+c)^2+9bc+4(b-c)^2}-\frac{1}{6(a+b+c)}\right)=$ $=\sum_{cyc}\frac{5a^2-5b^2-5c^2+4ab+4ac-3bc}{6(1+4b^2+bc+4c^2)}=\sum_{cyc}\frac{(a-b)(5a+10b+3c)-(c-a)(5a+10c+3b)}{12(1+4b^2+bc+4c^2)}=$ $=\sum_{cyc}(a-b)\left(\frac{5a+10b+3c}{12(1+4b^2+bc+4c^2)}-\frac{5b+10a+3c}{12(1+4a^2+ac+4c^2)}\right)=$ $=\sum_{cyc}\frac{(a-b)^2(15a^2+15b^2-22c^2+50ab+7ac+7bc)(1+4a^2+ab+4b^2)}{12\prod\limits_{cyc}(1+4a^2+ab+4b^2)}\geq$ $\geq\frac{(a-c)^2(15a^2+15c^2-22b^2+50ac+7ab+7bc)(1+4a^2+ac+4c^2)}{12\prod\limits_{cyc}(1+4a^2+ab+4b^2)}+$ $+\frac{(b-c)^2(15b^2+15c^2-22a^2+50bc+7ac+7ab)(1+4b^2+bc+4c^2)}{12\prod\limits_{cyc}(1+4a^2+ab+4b^2)}\geq$ $\geq\frac{(b-c)^2(15a^2-22b^2+7ab)(1+4a^2+ac+4c^2)}{12\prod\limits_{cyc}(1+4a^2+ab+4b^2)}+$ $+\frac{(b-c)^2(15b^2-22a^2+7ab)(1+4b^2+bc+4c^2)}{12\prod\limits_{cyc}(1+4a^2+ab+4b^2)}=$ $=\frac{(b-c)^2(a-b)^2(53a^2+53b^2-35c^2+74ab+ac+bc)}{12\prod\limits_{cyc}(1+4a^2+ab+4b^2)}\geq0$.

C-S is better: $\sum_{cyc}\frac{a}{1+9bc+4(b-c)^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a+9abc+4a(b-c)^2)}$ and the rest is Schur.

kunny wrote: Find the maximum value of real number $k$ such that \[\frac{a}{1+9bc+k(b-c)^2}+\frac{b}{1+9ca+k(c-a)^2}+\frac{c}{1+9ab+k(a-b)^2}\geq \frac{1}{2}\] holds for all non-negative real numbers $a,\ b,\ c$ satisfying $a+b+c=1$.

it is also solvable by cauchy schwarz. We obtain after using cauchy schwarz inequality that: $LHS \ge \frac{(a+b+c)^2}{a+b+c++27abc+\sum 4a(b-c)^2}$ then we have to prove that: $1 \ge 3abc+\sum 4ab(a+b)=3abc+\sum 4ab(1-c)$ <=> $1+9abc \ge 4ab+4bc+4ca$ <=> $(a+b+c)^3+9abc \ge 4(ab+bc+ca)(a+b+c)$ <=> $a^3+b^3+c^3+3abc \ge \sum ab(a+b)$ and done!

Generalization Let $a,\ b,\ c$ be non-negative real numbers satisfying $a+b+c=1$,and $\lambda \ge 0, 0\le k \le\frac{4}{9} \lambda $ . Prove that\[\frac{a}{1+\lambda bc+k(b-c)^2}+\frac{b}{1+\lambda ca+k(c-a)^2}+\frac{c}{1+\lambda ab+k(a-b)^2}\geq \frac{9}{9+\lambda}\]

Found it easy for problem 5 Taking $a=0, b=\frac{1}{2}, c=\frac{1}{2}$, we get that maximum value of $k$ is $4$. It remains to show that $\Sigma\frac{a}{1+bc+4b^2+4c^2} \ge \frac{1}{2}$. Using Titu's lemma we obtain that $$\Sigma\frac{a}{1+bc+4b^2+4c^2} \ge \frac{(a+b+c)^2}{1+3abc+4\Sigma ab(a+b)}=\frac{(a+b+c)^3}{1+3abc+4\Sigma ab(a+b)}$$. We will prove that $\frac{(a+b+c)^3}{1+3abc+4\Sigma ab(a+b)}\ge\frac{1}{2}$. We can rewrite as: $$2\Sigma a^3 + 6\Sigma ab(a+b)+ 12abc \ge 1+3abc+4\Sigma ab(a+b) \quad(1) $$Since $a+b+c=1$, then $\Sigma a^3 + 3\Sigma ab(a+b)+ 6abc=1$. As a result $(1)$ becomes: $$2\Sigma a^3 + 6\Sigma ab(a+b)+ 12abc \ge \Sigma a^3 + 7\Sigma ab(a+b)+ 9abc$$, which is equivalent to: $$\Sigma a^3+ 3abc \ge \Sigma ab(a+b) $$The last inequality is Schur's inequality of degree $1$, so we are done.

For collection \[\frac{a(a+b+c)}{(a+b+c)^2+9bc+4(b-c)^2} \geqslant \frac{8a^2+2a(b+c)-(b-c)^2}{12(a^2+b^2+c^2+ab+bc+ca)}.\]

Kunihiko_Chikaya wrote: Find the maximum value of real number $k$ such that \[\frac{a}{1+9bc+k(b-c)^2}+\frac{b}{1+9ca+k(c-a)^2}+\frac{c}{1+9ab+k(a-b)^2}\geq \frac{1}{2}\]holds for all non-negative real numbers $a,\ b,\ c$ satisfying $a+b+c=1$. See also https://artofproblemsolving.com/community/q1h1966069p13649308.

$\clubsuit \color{red}{\textit{\textbf{Proof:}}}$ Let $a=b=\frac{1}{2}$ and $c=0$, we get $k\le 4$. Notice that since $f(x)=\frac{1}{x}$ is a convex function, we have by Weighted Jensen's Inequality, \[\sum_{cyc} \frac{a}{1+9bc+4(b-c)^2}\ge \frac{1}{4\sum_{sym} ab^2+3abc + \sum_{cyc} a}\]we need to show that the $RHS \ge \frac{1}{2}$. This is true iff, \begin{align*} &2\ge 4\sum_{sym} ab^2+3abc + \sum_{cyc} a \\ \iff &(a+b+c)^3=1\ge 4\sum_{sym} ab^2+3abc \\ \iff &a^3+b^3+c^3+3abc\ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2 \end{align*}which is just Schur's 3rd degree Inequality. So, $\max (k)=4. \quad \blacksquare$

here is mine solution if something wrong please let me know and i m posting for storage My Solution:

Generalization 2 Find the maximum value of real number $k$.such that $$\dfrac{a}{\lambda -1 +9bc+k\left(b-c\right)^2}+\dfrac{b}{\lambda-1+9ca+k\left(c-a\right)^2}+\dfrac{c}{\lambda-1+9ab+k\left(a-b\right)^2}\geq \dfrac{1}{\lambda}$$ holds for all nonnegative reals $a,b,c$ satysfing $a+b+c=1$.