Fractions

Rearrange the given conditions to get $ab(a+b+1) = a^{2} +b^{2}$

Using cauchy-schwarz inequality, $a^{2}+b^{2}=ab(a+b+1) \geq \dfrac{a^{2} +b^{2} +2ab}{2}$

$\implies 2ab(a+b) \geq a^{2} +b^{2}$

Again by Cauchy Schwartz inequality $2ab(a+b) \geq a^{2} +b^{2}\geq \dfrac{(a+b)^{2}}{2}$

$\implies \boxed{4\geq \dfrac{a+b}{ab}}$

Equality holds when $a=b=\frac{1}{2}$

Consider $a+b=\eta$ & $ab=\omega$ .

Using few basic algebraic identities one can easily arrange above given relation as

$\omega=\eta^{2}/(3+\eta)$ .

Therefore the above

$1/a + 1/b$ = $1+ 3/\eta$

Also we have boundary conditions on $\eta$ , through :

$\eta = a/b + b/a -1$

Therefore boundries for $\eta$ are 1 & -3 .

Putting those we get answer as $\boxed{\boxed{4}}$