I'm having trouble showing the following.
If $a>b\ge 0$, then $$\frac{a+b}{2}-\sqrt{ab}<1 \iff \begin{cases}a+b<2, \text{ or,} \\ a+b \ge 2 \text{ and } (a-b)^2 < 4(a+b)-4.\end{cases}.$$
If $a+b<2$, the left-hand side holds trivially. I'm having trouble with the bottom line on the right-hand side.
Here are some attempts. \begin{align*} (a-b)^2 &< 4(a+b)-4\\ (a+b)^2 &< 4(a+b)-4+4ab\\ (a+b)(a+b-4) &< 4ab-4\\ &? \end{align*}
And in the other direction, \begin{align*} a+b &< 2(1+\sqrt{ab})\\ (a+b)^2 &< 4(1+2\sqrt{ab}+ab)\\ (a-b)^2 &< 4(1+2\sqrt{ab}) \le 4+4(a+b) & \text{GM-AM}\\ \end{align*} [so close!]
Any hints would be appreciated. Thanks!