Why is $ab+\frac{1}{2} - \frac{a+b}{2} >0$ for $0I am trying to show that the function $$ab+\frac{1}{2}-\frac{a+b}{2}$$ is positive when $0<a,b<1.$
Here is what I have done.

*

*The arithmetic-geometric mean inequality doesn't seem like the way to go because it implies $$ab < \sqrt{ab} \le \frac{a+b}{2}.$$
This implies $$ab+\frac{1}{2}-\frac{a+b}{2}\le \frac{1}{2}.$$


*When I fix values for $a$ (or $b$) and sketch traces of the function $ab+\frac{1}{2}-\frac{a+b}{2}$, I obtain positive values on the graph.

 A: $ab+\dfrac12-\dfrac{a+b}2=\dfrac{ab}2+\dfrac{\color{red}{ab}\color{blue}{+1-a}\color{red}{-b}}2=$
$=\dfrac{ab}2+\dfrac{\color{blue}{(1-a)}\color{red}{-b(1-a)}}2=$
$=\dfrac{ab}2+\dfrac{(1-a)(1-b)}2>0\;,\;$ since $\;0<a,\,b<1\,.$
A: The function $f(a, b) = ab+\frac{1}{2}-\frac{a+b}{2}$ is linear in both arguments, and a linear function attains its minimum (and maximum) on any closed interval at one of the endpoints.
It follows that for $0 < a < 1$ and $0 < b < 1$
$$
 f(a, b) \ge \min\bigl(f(a, 0), f(a, 1)\bigr) = \min\bigl(\frac{1-a}{2}, \frac a2\bigr) > 0 \, .
$$
A: As a rule, when you are dealing with numbers in $[0,1]$, you can write them as $a = \frac{u}{u+v}$ and $b= \frac{p}{p+q}$, where $u$, $v$, $p$, $q$ are positive.  Substituting we get
$$a b + \frac{1}{2} - \frac{a+b}{2} = \frac{p u + q v}{2 (p + q) (u + v)}$$
clearly positive. We can substitute back if we wish, $u+v= p+q=1$, $u = a$, $p=b$ and get the expression
$$\frac{ a b + (1-a)(1-b)}{2}$$
A: You have shown that
$$
f(a, b) = ab+\frac{1}{2}-\frac{a+b}{2} < \frac{1}{2}
$$
for $0 < a, b < 1$, because
$$
ab < \sqrt{ab} \le \frac{a+b}{2}.
$$
But $ f(a, b) + f(1-a, b) = \frac 12$,
and therefore your upper bound also gives the desired lower bound:
$$
 f(a, b) = \frac 12 - \underbrace{f(1-a, b)}_{<1/2} > 0 \, .
$$
