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.

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 $c = \frac12$.
