Show $\sqrt{x (1-y)} - \sqrt{y (1-x)} + y\sqrt{x} - x\sqrt{y} - \frac{x - y}{3} > 0$

Question

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?

Answer 1

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: $$\frac 13+s^2-t^2<\frac 13+1-\frac 12 = \frac 56 <1<\frac st.$$

Answer 2

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.