Inequality, $\frac{u+v}{2}-\frac{1}{4}$ versus $uv$ 
On the unit square, where is $f(u,v) = uv$ greater than $g(u,v) = \frac{u+v}{2}-\frac{1}{4}$?

I know that

*

*$f \left( \frac{1}{2} ,  \frac{1}{2}\right ) =  \frac{1}{4} = g \left( \frac{1}{2} ,  \frac{1}{2}\right ) $
From playing with numbers,

*

*on $\left(0, \frac{1}{2}\right) \times \left(0, \frac{1}{2}\right)$ and $\left(\frac{1}{2}, 1\right) \times \left(\frac{1}{2}, 1\right)$, $f \geq g$

*and $g\leq f$ elsewhere.

How do I show this mathematically?
On $(0, \frac{1}{2}) \times (0, \frac{1}{2})$, is it a valid argument to notice that the partial derivative of $f$ are less than $\frac{1}{2}$ while the partial derivatives of $g$ are both equal to $\frac{1}{2}$, and conclude that $f$ decreases "slower" and is thus larger valued that $g$ (since they are equal at $\left( \frac{1}{2} ,  \frac{1}{2} \right)$)? (and similarly, on $\left(\frac{1}{2}, 1\right) \times \left(\frac{1}{2}, 1\right)$, $f$ increases "faster" and is thus greater than $g$)?
I have no idea what to do on $\left(\frac{1}{2}, 1\right) \times \left(0, \frac{1}{2}\right)$ and $\left(0, \frac{1}{2}\right) \times \left(\frac{1}{2}, 1\right) $.

$$\begin{align}
f(u, v) - g(u, v) &= uv - \left(\frac{u+v}{2} - \frac14\right) \\
 &= uv - \frac u2 - \frac v2 + \frac 14 \\
 &= \left(u - \frac12\right)\left(v - \frac12\right)
\end{align}$$
