Proof that $a+b+\frac{1}{1+\sqrt{ab}}\geq\frac{5}{2}$

We have: $4(ab)^2 = (a+b)^2 \ge 4ab \implies ab \ge 1\implies LHS \ge 2\sqrt{ab} + \dfrac{1}{1+\sqrt{ab}}= f(t), t = \sqrt{ab} \ge 1\implies f'(t) = 2 - \dfrac{1}{(1+t)^2}> 0\implies f(t) \ge f(1) = 2+\dfrac{1}{2} = \dfrac{5}{2} \implies LHS \ge \dfrac{5}{2} = RHS $. Equality occurs when $a = b = 1$.