Thai Mathematical Olympiad

Solution. To prove the second inequality, we may assume $x=\max \{x,y,z\}$. Then we get $$\frac{4z^2+x^2}{z^2+4x^2}\le 1,\frac{4x^2+y^2}{x^2+4y^2}<4,\text{and}\frac{4y^2+z^2}{y^2+4z^2}<4.$$ So the desired inequality follows.

Next we prove the first inequality. By the AM-GM inequality, we have $4x{y}^2\le y^3+4x^{2}y$. Then $$\begin{array}{rl}{y}^3+4x^{2}y+3x^3& >y^3+4x^{2}y\ge 4x{y}^2>3x{y}^2\\ ⟺y^3+4x^{2}y+4x^3+x{y}^2& >4x{y}^2+x^3\\ ⟺\frac{4x^2+y^2}{x^2+4y^2}& >\frac{x}{x+y}.\end{array}$$ Hence $$\sum _{cyc}\frac{x}{x+y}<\sum _{cyc}\frac{4x^2+y^2}{x^2+4y^2}.$$ Now apply the Cauchy-Schwartz inequality, we have $$\sum _{cyc}\frac{x}{\sqrt{2(x^2+y^2)}}=\sum _{cyc}\frac{x}{\sqrt{(1+1)(x^2+y^2)}}\le \sum _{cyc}\frac{x}{x+y}<\sum _{cyc}\frac{4x^2+y^2}{x^2+4y^2},$$ which is the first inequality. $\square$