Prove $ \min \left(a+b+\frac1a+\frac1b \right) = 3\sqrt{2}\:$ given $a^2+b^2=1$

Notice $$\begin{align} a + b + \frac1a + \frac1b = &\; \left(a + \frac{1}{2a} + \frac{1}{2a}\right) + \left(b + \frac{1}{2b} + \frac{1}{2b}\right)\\ \\ \color{blue}{\rm AM \ge \rm GM \rightarrow\quad} \ge &\; 3\left[\left(\frac{1}{4a}\right)^{1/3} + \left(\frac{1}{4b}\right)^{1/3}\right]\\ \color{blue}{\rm AM \ge \rm GM \rightarrow\quad} \ge &\; 6 \left(\frac{1}{16ab}\right)^{1/6}\\ \color{blue}{a^2 + b^2 \ge 2 ab \rightarrow\quad} \ge &\; 6 \left(\frac{1}{8(a^2+b^2)}\right)^{1/6}\\ = &\; 6 \left(\frac18\right)^{1/6} = \frac{6}{\sqrt{2}} = 3\sqrt{2} \end{align} $$ Since the value $3\sqrt{2}$ is achieved at $a = b = \frac{1}{\sqrt{2}}$, we have

$$\min \left\{ a + b + \frac1a + \frac1b : a, b > 0, a^2+b^2 = 1 \right\} = 3\sqrt{2}$$

Notes

About the question how do I come up with this. I basically know the minimum is achieved at $a = b = \frac{1}{\sqrt{2}}$. Since the bound $3\sqrt{2}$ on RHS of is optimal, if we ever want to prove the inequality, we need to use something that is tight when $a = b$. If we want to use AM $\ge$ GM, we need to arrange the pieces so that all terms are equal. That's why I split $\frac1a$ into $\frac{1}{2a} + \frac{1}{2a}$ and $\frac1b$ into $\frac{1}{2b} + \frac{1}{2b}$ and see what happens. It just turns out that works.

My answer is a little roundabout but without calculus and without pictures or symmetry:

Arithmetic-geometric inequality: $$ a+b \geq 2\sqrt{ab} $$ Harmonic-geometric inequality and some rearrangement: $$ \sqrt{ab} \geq \frac{2}{\frac{1}{a}+\frac{1}{b}}$$ $$ \frac{1}{a} + \frac{1}{b}\geq \frac{2}{\sqrt{ab}}$$ Add both results to get $$ a+b+\frac{1}{a}+\frac{1}{b} \geq2\left(\sqrt{ab}+\frac{1}{\sqrt{ab}}\right)~~~~~~~~~~(1)$$ Also note that by the given constraint $$ 0\leq(a-b)^2 = a^2+b^2-2ab = 1 - 2ab$$ and therefore $$ \sqrt{ab} \leq \frac{1}{2}\sqrt{2}$$ Now let $x:=\sqrt{ab}$.

So $0\leq x \leq \frac{1}{2}\sqrt{2} < 1$

To finish off, all we need to show is that the right-hand side $2(x+\frac{1}{x})$ of inequality $(1)$ is minimal if $x$ is maximal and therefore $\frac{1}{2}\sqrt{2}$ because then $$ a+b+\frac{1}{a}+\frac{1}{b} \geq2\left(\frac{1}{2}\sqrt{2}+\sqrt{2}\right) = 3 \sqrt{2}$$ is always true.

To show that, let's show that the function $f:x\mapsto x+\frac{1}{x}$ is decreasing on the interval $(0,1)$. That's easy with calculus. Without:

Let's choose $h,x$ arbitrarily such that $0<h<1$ and $0<x<x+h<1$.

Then rearrange equivalently or backwards-implicatively to get to our monotonicity claim from a true statement

$$x+h+\frac{1}{x+h} < x+\frac{1}{x}$$ $$ h+\frac{1}{x+h} < \frac{1}{x}$$ Multiply through $$ hx(x+h) + x < x + h$$ And since $h+x < 1$: $$\Leftarrow hx + x \leq x +h $$ Since also $x<1$: $$\Leftarrow h + x \leq h+x $$ which is true. Since $x,h$ were arbitrary from $(0,1)$, this proves monotonicity and hence the claim.

We will square the whole inequality $$\frac{(a+b)^2}{(ab)^2}+(a+b)^2+2\frac{(a+b)^2}{ab}\geq 18$$ simplifying and using that $$a^2+b^2=1$$ we get

$$2(ab)^3-13(ab)^2+4ab+1\geq0$$ this is equivalent to $$(2ab-1)((ab)^2-6ab-1)\geq 0$$

Now we have $$a^2+b^2\geq 2ab$$ this is $$ab\le \frac{1}{2}$$

so both factors $$2ab-1,(ab)^2-6ab-1$$ are non posivite, thus their product is non negative.