Estimates for $\sqrt{x(1-x)}$ Assume $x,y\in[0,1]$ and $|x-y|<\frac{1}{4}$
Any idea how to see that these inequalities hold:
$|\sqrt{x(1-x)}-\sqrt{y(1-y)}|^2<|\sqrt{x}-\sqrt{y}|$ for $x,y\in[\frac{1}{4},1]$
and
$|\sqrt{x(1-x)}-\sqrt{y(1-y)}|^2<|\sqrt{1-x}-\sqrt{1-y}|$ for $x,y\in[0,\frac{3}{4}]$
I already tried some simple calculations but couldn't see it. Maybe there is a simple trick?
 A: The first inequality is untrue in general as it fails when $x,y$ both approach $1$. For example, take $(x,y)=(0.999,0.9)$ which satisfies $x,y\in[\frac14,1]$ and $|x-y|=0.099<\frac14$ but gives
$$\left|\sqrt{x\left(1-x\right)}-\sqrt{y\left(1-y\right)}\right|^{2}=0.0720\nless0.0508=\left|\sqrt{x}-\sqrt{y}\right|$$
As has been established in the comments, the second inequality is equivalent to the first under the transformation $(x',y')=(1-x,1-y)$, so neither inequality is generally true.
To assess where the first inequality does hold, consider the sign of the function defined by $$f(x,y)=\underbrace{\left|\sqrt{x}-\sqrt{y}\right|}_{\text{RHS}}-\underbrace{\left|\sqrt{x\left(1-x\right)}-\sqrt{y\left(1-y\right)}\right|^{2}}_{\text{LHS}}$$
$$=\left[\sqrt{x}+2\sqrt{x\left(1-x\right)}\sqrt{y\left(1-y\right)}\right]-\left[\sqrt{y}+x\left(1-x\right)+y\left(1-y\right)\right]$$
where we dispense with the absolute value signs by expanding the square via $|z|^2=z^2$ and assume WLOG that $x>y$. The function $f$ is continuous, so solving the inequality amounts to finding its zeros and determining its sign near them. Since $f$ is algebraic, this is equivalent to solving the polynomial given by clearing its radicals.
