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

If $a=b=\frac{1}{\sqrt2}$ then $a+b+\frac{1}{ab}=\sqrt2+2$.
We'll prove that it's a minimal value.
Indeed, let $a+b=2u$ and $ab=v^2$.
Hence, the condition gives $4u^2-2v^2=1$,
which says that $v^2=\frac{4u^2-1}{2}$ and $4u^2=1+2v^2\leq1+2u^2$, which gives $u\leq\frac{1}{\sqrt2}$
and we need to prove that
$$2u+\frac{2}{4u^2-1}\geq\sqrt2+2$$ or
$$(1-\sqrt2u)(2\sqrt2-1+4u+2(1-2u^2))\geq0$$
and we are done!
