If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$ If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$
This can be easily done by calculas but is there any way to do do this by algebra

$$ a + b + \frac{1}{ab} = a + b + \frac{ a^2 + b^2}{ab} \geq a + b + 2 $$
I have used AM-GM ineq:
$$ \frac{a^2 + b^2}{2} \geq ab $$
Remark: IT is still left to show that $a+b \geq \sqrt{2} $ constrained to $a^2 + b^2 = 1 $. See   A Blumenthal's solution.
