Thai Mathematical Olympiad

Without loss of generality, we may assume that $a\ge b\ge c$. Then $$\frac{(c-a)(c-b)}{3(c+ab)}\ge 0$$ and $$\frac{(a-b)(a-c)}{3(a+bc)}+\frac{(b-a)(b-c)}{3(b+ca)}=\frac{c(a-b{)}^2}{3}\left(\frac{1+a+b-c}{(a+bc)(b+ca)}\right)\ge 0$$ Therefore $$\sum _{\text{cyc}}\frac{(a-b)(a-c)}{3(a+bc)}\ge 0$$

$$\begin{array}{rl}\sum _{\text{cyc}}\frac{1}{a+bc}& \le \frac{9}{2(ab+bc+ca)}\\ \iff \sum _{\text{cyc}}\frac{1}{a(a+b+c)+3bc}& \le \frac{3}{2(ab+bc+ca)}\\ \iff \sum _{\text{cyc}}\left[\frac{1}{2(ab+bc+ca)}-\frac{1}{a(a+b+c)+3bc}\right]& \ge 0\\ \iff \sum _{\text{cyc}}\frac{(a-b)(a-c)}{a(a+b+c)+3bc}& =\sum _{\text{cyc}}\frac{(a-b)(a-c)}{3(a+bc)}\ge 0.\end{array}$$ By the AM-GM inequality, we have $$\frac{1}{a\sqrt{2(a^2+bc)}}=\frac{\sqrt{b+c}}{\sqrt{2a}\sqrt{(ab+ac)(a^2+bc)}}\ge \frac{\sqrt{2(b+c)}}{\sqrt{a(a+b)(a+c)}}$$ So it suffices to prove that $$\sum _{\text{cyc}}\sqrt{\frac{b+c}{2a}}\cdot \frac{1}{(a+b)(a+c)}\ge \frac{9}{4(ab+bc+ca)}$$ Since $\sqrt{\frac{b+c}{2a}}\le \sqrt{\frac{c+a}{2b}}\le \sqrt{\frac{a+b}{2c}}$ and $$\frac{1}{(a+b)(a+c)}\le \frac{1}{(b+c)(b+a)}\le \frac{1}{(c+a)(c+b)},$$ by Chebyshev's inequality, we get $$\begin{array}{rl}\sum _{\text{cyc}}\sqrt{\frac{b+c}{2a}}\cdot \frac{1}{(a+b)(a+c)}& \ge \frac{1}{3}\left(\sum _{\text{cyc}}\sqrt{\frac{b+c}{2a}}\right)\left(\sum _{\text{cyc}}\frac{1}{(a+b)(a+c)}\right)\\ & =\frac{2}{(a+b)(b+c)(c+a)}\left(\sum _{\text{cyc}}\sqrt{\frac{b+c}{2a}}\right).\end{array}$$ In view of these estimates, we now see that it suffices to show that $$\sum _{\text{cyc}}\sqrt{\frac{b+c}{2a}}\ge \frac{9(a+b)(b+c)(c+a)}{8(ab+bc+ca)}.$$ Let $t:=\sqrt[6]{\frac{(a+b)(b+c)(c+a)}{8abc}}$. Clearly $t\ge 1$. Using the AM-GM inequality, we find that $$\sum _{\text{cyc}}\sqrt{\frac{b+c}{2a}}\ge 3t.$$

It is easy to verify that $\frac{9(a+b)(b+c)(c+a)}{8(ab+bc+ca)}=\frac{27t^6}{8t^6+1}$; so, it is enough to prove that $3t\ge \frac{27t^6}{8t^6+1}$ or $8t^6-9t^5+1\ge 0$. Since $t\ge 1$, this inequality is true and the proof is completed. $\square$