Minimal c satisfying $x+y-(xy)^c \geq 0$ for all $x,y\in [0,1]$ What is the minimal real $c$ satisfying $x+y-(xy)^c \geq 0$ for all $x,y \in [0,1]$?
Experimentally (though my experiments weren't necessarily accurate enough) I reached as low as $c=\tfrac{13}{32}$, where $c=\tfrac{12}{32}$ violates it.
 A: Your minimal value also does violate it. Consider $x=y=0.02$, then
$$x+y-(xy)^{13/32}=-0.00164730...$$
Consider the implicit function $$x+y=(xy)^c$$
It is easy to show that the plot is symmetric over $y=x$. Changing coordinates to polar and considering function $r(\theta)$ we can show that $\theta=\tfrac\pi4$ is the maximum (see below). Your inequality then is equivalent to $r\geqslant r(\theta)$. Consider the path $x=y$ and do the substitution $x=y=t$ to get
$$2t=t^{2c}$$
Variable $t$ is then the maximum of $r(\theta)$. Find the function $t(c)$:
$$2=t^{2c-1}$$
$$t=2^{1/(2c-1)}$$
We can use hyperreal numbers to find its zero
$$2^{1/(2c-1)}=0$$
$$\frac{1}{(2c-1)}=-\infty$$
$$2c-1=0^-$$
$$c=\frac12^-$$
so the minimal $c$ is $$c=\frac12$$
Function $r(\theta)$
For $x+y=(xy)^c$ we substitute $x=r\cos(\theta)$, $y=r\sin(\theta)$ to get
$$r(\sin\theta+\cos\theta)=(r^2\sin\theta\cos\theta)^c$$
$$\sin\theta+\cos\theta=(r^{2-1/c}\sin\theta\cos\theta)^c$$
$$(\sin\theta+\cos\theta)^{1/c}=r^{2-1/c}\sin\theta\cos\theta$$
$$r^{2-1/c}=\frac{(\sin\theta+\cos\theta)^{1/c}}{\sin\theta\cos\theta}$$
$$r=\left(
\frac{(\sin\theta+\cos\theta)^{1/c}}{\sin\theta\cos\theta}
\right)^{c/(2c-1)}$$
Its derivative is equal to
$$r'=\frac{c}{2c-1}\cdot
\\
\left(
\frac{(\sin\theta+\cos\theta)^{1/c}}{\sin\theta\cos\theta}
\right)^{(-c-1)/(2c-1)}\cdot
\\
\left(
\frac{
\frac1c(\sin\theta+\cos\theta)^{1/c-1}(\cos\theta-\sin\theta)\sin\theta\cos\theta-(\sin\theta+\cos\theta)^{1/c}(
\cos\theta\cos\theta-\sin\theta\sin\theta
)
}{
(\sin\theta\cos\theta)^2
}
\right)$$
And is equal to $0$ when 
$$\left(
\frac{(\sin\theta+\cos\theta)^{1/c}}{\sin\theta\cos\theta}
\right)^{(-c-1)/(2c-1)}=0$$
$$\sin\theta+\cos\theta=0$$
$$\theta=\pi n-\tfrac\pi4$$
None of them belong to $[0;\pi/2]$, second one:
$$(\sin\theta+\cos\theta)^{-1}(\cos\theta-\sin\theta)\sin\theta\cos\theta=c(
\cos\theta\cos\theta-\sin\theta\sin\theta
)$$
$$(\cos\theta-\sin\theta)\sin\theta\cos\theta=c\cos(2\theta)(\sin\theta+\cos\theta)$$
Plug $\theta=\tfrac\pi4$ to get
$$\cos\theta=\sin\theta,\cos(2\theta)=0 \implies 0=0$$
A: Your answer is $c = \frac12$.
If $c = 1/2$,
$\begin{array}\\
x+y-(xy)^c
&=x+y-(xy)^{1/2}\\
&=x-(xy)^{1/2}+y/4-y/4+y\\
&=(\sqrt{x}-\frac12\sqrt{y})^2-y/4+y\\
&=(\sqrt{x}-\frac12\sqrt{y})^2+3y/4\\
&\ge 0\\
\end{array}
$
so it is true for
$c \ge \frac12$
(since
$(xy)^{1/2+d}
\le (xy)^{1/2}
$
for
$d \ge 0$).
Suppose $c = \frac12-d$
where
$\frac12 > d > 0$.
Let $y = x$.
Then
$\begin{array}\\
x+y-(xy)^c
&=2x-(x^2)^{1/2-d}\\
&=2x-x^{1-2d}\\
&=x(2-x^{-2d})\\
\end{array}
$
and this is negative when
$2 < x^{-2d}$
or
$x^{2d} < \frac12$
or
$x <\frac1{2^{1/(2d)}}
$.
Therefore,
if $c < \frac12$,
there is $x$ and $y$
for which the inequality is false.
Therefore,
your answer is $c = \frac12$.
