16th Junior Turkish Mathematical Olympiad

The Cauchy-Schwarz inequality implies that $(x+y)(x^3+y^3)\ge (x^2+y^2{)}^2$. Therefore, $$1\le \frac{(x+y)(x^3+y^3)}{(x^2+y^2{)}^2}.$$ As $0\le ((x-y{)}^2-2xy{)}^2$, we have $$4xy(x-y{)}^2\le (x-y{)}^4+4x^2{y}^2.$$ $$\Rightarrow 8x^{3}y+8y^{3}x\le x^4+18x^2{y}^2+y^4.\Rightarrow 8x^{3}y+8y^{3}x+8x^4+8y^4\le 9x^4+18x^2{y}^2+9y^4.$$ $$\Rightarrow 8(x+y)(x^3+y^3)\le 9(x^2+y^2{)}^2.\text{ Thus,}$$ $$\frac{(x+y)(x^3+y^3)}{(x^2+y^2{)}^2}\le \frac{9}{8}.$$