<s>[INST] Show $\sqrt{x (1-y)} - \sqrt{y (1-x)} + y\sqrt{x} - x\sqrt{y} - \frac{x - y}{3} > 0$ Consider the inequality 
$$
\sqrt{x (1-y)} - \sqrt{y (1-x)} + y\sqrt{x} -  x\sqrt{y} - \frac{x - y}{3} > 0
$$
where $0<y<x<1$ and $x+y\leq 1$.
One can, for instance, consider optimizing the expression using the Karush-Kuhn-Tucker conditions. However, the computations involved are tedious and unwieldy. 
Can the inequality be shown in a simpler way?
 [/INST] For a traditional kind of proof I use beside $\,0<y<x<1\,$ and $\,x+y\leq 1\,$ also 
first derivation ($:=0$) and second derivation ($\neq 0$) of a term to get minimum and maximum. 
Equivalent to the given inequation is $\enspace\displaystyle \sqrt{x(1-y)}-\sqrt{y(1-x)} > x\sqrt{y}- y\sqrt{x} + \frac{x-y}{3}$ 
and we devide by $\,x-y\,$ to get $\enspace\displaystyle \frac{1}{\sqrt{x(1-y)}+\sqrt{y(1-x)}} > \frac{1}{\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}}} + \frac{1}{3} \,$ .
Using first and second derivation we get 
$\displaystyle \sqrt{x(1-y)}+\sqrt{y(1-x)} \leq \max(\sqrt{x(1-y)}+\sqrt{y(1-x)}) = 1 \enspace$ and 
$\displaystyle \frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}} \geq \frac{1}{\sqrt{1-x}}+\frac{1}{\sqrt{x}} \geq \min(\frac{1}{\sqrt{1-x}}+\frac{1}{\sqrt{x}}) = 2\sqrt{2}\enspace$ so that follows 
$\displaystyle \frac{1}{\sqrt{x(1-y)}+\sqrt{y(1-x)}} \geq 1 > \frac{1}{2\sqrt{2}} + \frac{1}{3} \geq \frac{1}{\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}}} + \frac{1}{3} \enspace$ as wished.
</s>