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?

If you let $x = \sin^2 a, y = \sin^2 b$ with $0<b<a<\pi/2$, then your inequality is equivalent to:
$$\sin a\cos b - \sin b\cos a +\sin^2b\sin a - \sin^2a\sin b >\dfrac{\sin^2 a-\sin^2 b}{3}$$ or $$\sin(a-b)>(\sin a-\sin b)\left(\frac 13+\sin a\sin b\right)$$
or $$\dfrac{\cos\left(\frac {a-b}2\right)}{\cos\left(\frac{a+b}{2}\right)}>\frac{1}{3}+\frac 12(\cos(a-b)-\cos(a+b)).$$
Here, the remaining requirement is the same as $\sin a\leq\cos b = \sin(\pi/2-b)$, which is the same as: $$a+b\leq\dfrac{\pi}{2}.$$
Now let $1>\cos\left(\frac{a-b}{2}\right) = s>\cos\left(\frac{a+b}{2}\right)=t\geq\dfrac{1}{\sqrt{2}}.$ Then, the equivalent inequality is:
$$\frac st>\frac 13+\frac 12(2s^2-1 - 2t^2+1) = \frac 13+s^2-t^2$$
But this one is trivial as:
$$\frac 13+s^2-t^2<\frac 13+1-\frac 12 = \frac 56 <1<\frac st.$$ 
