<s>[INST] 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.

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*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.

 [/INST] 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
$$\frac{ a b + (1-a)(1-b)}{2}$$
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