<s>[INST] 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
 [/INST] 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 $a + b \geq \sqrt{2}$.
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