Thailand Mathematical Olympiad

First we prove that $$\frac{a^2}{\sqrt{a^4+4b+4c}}+\frac{b^2}{\sqrt{b^4+4c+4a}}+\frac{c^2}{\sqrt{c^4+4a+4b}}\ge 1$$ Applying AM-GM inequality, we conclude that $$\begin{array}{rl}\frac{a^2}{\sqrt{a^4+4b+4c}}& =\frac{a^2}{\sqrt{a^4+4a{b}^{2}c+4ab{c}^2}}\\ & \ge \frac{a^2}{\sqrt{a^4+(a^2{b}^2+a^2{c}^2+b^2{c}^2+b^4)+(a^2{b}^2+a^2{c}^2+b^2{c}^2+c^4)}}\\ & =\frac{a^2}{a^2+b^2+c^2}\end{array}$$ Similarly, $$\frac{b^2}{\sqrt{b^4+4c+4a}}\ge \frac{b^2}{b^2+c^2+a^2};\frac{c^2}{\sqrt{c^4+4a+4b}}\ge \frac{c^2}{c^2+a^2+b^2}$$ Therefore $$\frac{a^2}{\sqrt{a^4+4b+4c}}+\frac{b^2}{\sqrt{b^4+4c+4a}}+\frac{c^2}{\sqrt{c^4+4a+4b}}\ge 1$$

Now it remains to prove that $$\frac{1}{\sqrt{a^3+2b^3+6}}+\frac{1}{\sqrt{b^3+2c^3+6}}+\frac{1}{\sqrt{c^3+2a^3+6}}\le 1$$ Applying the AM-GM inequality we conclude that $$\begin{array}{rl}\frac{1}{\sqrt{a^3+2b^3+6}}& \le \frac{1}{\sqrt{3a{b}^2+6}}\\ & =\sqrt{\frac{abc}{3a{b}^2+6abc}}=\sqrt{\frac{1}{3}\left(\frac{c}{b+2c}\right)}\\ & \le \frac{1}{2}\left(\frac{1}{3}+\frac{c}{b+2c}\right)\\ & =\frac{1}{4}\left(\frac{5}{3}-\frac{a}{a+2b}\right).\end{array}$$

Therefore using the Cauchy-Schwartz inequality we have $$\begin{array}{rl}& \frac{1}{\sqrt{a^3+2b^3+6}}+\frac{1}{\sqrt{b^3+2c^3+6}}+\frac{1}{\sqrt{c^3+2a^3+6}}\\ \le & \frac{5}{4}-\frac{1}{4}\left(\frac{b}{b+2c}+\frac{c}{c+2a}+\frac{a}{a+2b}\right)\\ \le & \frac{5}{4}-\frac{(a+b+c{)}^2}{4(b(b+2c)+c(c+2a)+a(a+2b))}=1.\end{array}$$ Hence the proof of this problem follows. □