I got $$p\geq 1$$for $$a=1,b=2$$Sonnhard.

parmenides51 wrote: Find the smallest real constant $p$ for which the inequality holds $\sqrt{ab}- \frac{2ab}{a + b} \le p \left( \frac{a + b}{2} -\sqrt{ab}\right)$ with any positive real numbers $a, b$. Substitute $a=1, b=c^2\neq 1$ to get $p\geq\frac{2c}{c^2+1}$ We want $pc^2-2c+p\geq 0$ for all $c\in\mathbb{R}-\{1,-1\}$ So, the discriminant of this quadratic is non-negative, and also by $x\to\infty$, we have $p\geq 0$. $4-4p^2\leq 0\implies p\geq 1$ We prove this inequality for $p=1$ $\sqrt{ab}-\frac{2ab}{a+b}\leq \frac{a+b}{2}-\sqrt{ab}$ Let $a+b=x, ab=y^2$ The inequality is equivalent to $(x-2y)^2\geq 0$, which is true.