<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.

*

*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] 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:
$$
 f(a, b) = \frac 12 - \underbrace{f(1-a, b)}_{<1/2} > 0 \, .
$$
</s>