YadisBeles wrote: Let $x,y$ be positive real numbers such that $x+y=1$. Prove that$\frac{x}{x^2+y^3}+\frac{y}{x^3+y^2}\leq2(\frac{x}{x+y^2}+\frac{y}{x^2+y})$. Let $x=ty$. Hence, we need to prove that $\frac{2}{t^2+t+1}\geq\frac{t}{t^3+t^2+1}+\frac{1}{t^3+t+1}$, which is $(t-1)^2(t^4+2t^3+t^2+2t+1)\geq0$. Done!

YadisBeles wrote: Let $x,y$ be positive real numbers such that $x+y=1$. Prove that$\frac{x}{x^2+y^3}+\frac{y}{x^3+y^2}\leq2(\frac{x}{x+y^2}+\frac{y}{x^2+y})$. The inequality is equivalent to Let $x,y$ be positive real numbers such that $x+y=1$.Prove that$$\frac{x}{x^2+y^3}+\frac{y}{x^3+y^2}\leq \frac{1}{x+y^2}+\frac{1}{x^2+y}$$Sqing inequality chain (2)

YadisBeles wrote: Let $x,y$ be positive real numbers such that $x+y=1$. Prove that$\frac{x}{x^2+y^3}+\frac{y}{x^3+y^2}\leq2(\frac{x}{x+y^2}+\frac{y}{x^2+y})$. Another approach: First, it is not hard to show that $x+y^2=y+x^2=1-xy$ and $xy\leq\frac{1}{4}$ Next, by Cauchy: \begin{align*} \frac{x}{x^2+y^3}+\frac{y}{x^3+y^2}&=\frac{x(1+y)}{(x^2+y^3)(1+y)}+\frac{y(1+x)}{(x^3+y^2)(1+x)} \\ &\leq \frac{x(1+y)}{(x+y^2)^2}+\frac{y(1+x)}{(x^2+y)^2} \\ &= \frac{1+2xy}{(x+y^2)(1-xy)}\\ &\leq \frac{2}{x+y^2} \\ &=2(\frac{x}{x+y^2}+\frac{y}{x^2+y}) \end{align*}