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 $a>0$ and $b>0$, 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

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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 $a + b \geq \sqrt{2}$.

Another way is to use AM-GM while preserving the point of equality, i.e.:

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}

$$ a + b + \frac{1}{ab} = a + b + \frac{ a^2 + b^2}{ab} \geq a + b + 2 $$

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.

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
and we are done!

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 $x = \frac{\pi}4$.