Irska

Multiplying through by $\sqrt{ab}$ we see that we require the largest number $q$ so that $$\frac{a+b}{\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}p\ge q,\forall a,b>0.$$ To deal with this, let $$s=\frac{a+b}{\sqrt{ab}},$$ so that the inequality becomes $$s+\frac{p}{s}\ge q.$$ This suggests that we should look for the minimum value of the function $$f(s)=s+\frac{p}{s}.$$ But $$f(s)-2\sqrt{p}={\left(\sqrt{s}-\frac{\sqrt{p}}{\sqrt{s}}\right)}^2\ge 0,$$ with equality iff $s=\sqrt{p}$. Hence, $f(s)\ge 2\sqrt{p}$, $\forall s>0$, i.e., $q\le 2\sqrt{p}$. In other words, for all $a,b>0$, $$\frac{1}{a}+\frac{1}{b}+\frac{p}{a+b}\ge \frac{2\sqrt{p}}{\sqrt{ab}}.$$ To prove that $2\sqrt{p}$ is the best constant, we must confirm that there is a pair of positive numbers $a,b$ such that $$\frac{a+b}{\sqrt{ab}}=\sqrt{p}⟺t+\frac{1}{t}=\sqrt{p}.\text{with }t=\sqrt{\frac{a}{b}}.$$ But $p\ge 4$. Hence $(\sqrt{p}\pm \sqrt{p-4})/2$ are two positive solutions of the quadratic equation $t^2-\sqrt{p}t+1=0$. Choose $t$ to be any one of these, let $a=t^2,b=1$ and confirm that $$\frac{1}{t^2}+1+\frac{p}{t^2+1}=\frac{2\sqrt{p}}{t}$$ Hence $q=2\sqrt{p}$.