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: Another way is to use AM-GM while preserving the point of equality, i.e.: 
\begin{align}
a+b+\frac1{ab} &= a+b+\frac1{2 \sqrt2 ab} + \left(1-\frac1{2\sqrt2}\right)\frac1{ab} \\
&\geqslant \frac3{\sqrt2}+\left(1-\frac1{2\sqrt2}\right)\frac2{a^2+ b^2} \\
&= \frac3{\sqrt2}+\left(1-\frac1{2\sqrt2}\right)\cdot 2 = 2+\sqrt{2}
\end{align}
A: $$ a + b + \frac{1}{ab} = a + b + \frac{ a^2 + b^2}{ab} \geq a + b + 2 $$
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.
A: 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!
A: Let $a = \sin x, b = \cos x$. Then we need to find the minimum of the function $\cos x + \sin x + \sec(x) \csc(x)$ which is $2+ \sqrt{2}$ at $x = \frac{\pi}4$.
A: Edit: The last line is incorrect.
In light of @Lemur's answer, it will suffice to show that 
$$
a + b \geq \sqrt{2}
$$
where $a,b > 0, a^2 + b^2 = 1$ is enforced.
To see this, note that by AMGM, $ab \leq \frac{a^2 + b^2}{2} = \frac{1}{2}$, and so
$$
(a + b)^2 = a^2 + b^2 + 2 ab \geq 1 + 2 ab \geq 1 + 1 = 2
$$
Thus $a + b \geq \sqrt{2}$.
