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

Prove that

$$\min\left(a+b+\frac1a+\frac1b\right) = 3\sqrt{2}$$
Given $$a^2+b^2=1 \quad(a,b \in \mathbb R^+)$$
Without using calculus.
$\mathbf {My Attempt}$
I tried the AM-GM, but this gives $\min = 4$.

I used Cauchy-Schwarz to get $\quad (a+b)^2 \le 2(a^2+b^2) = 2\quad \Rightarrow\quad a+b\le \sqrt{2}$
But using Titu's Lemma I get $\quad \frac1a+\frac1b \ge \frac{4}{a+b}\quad \Rightarrow\quad \frac1a+\frac1b \ge 2\sqrt{2}$
I'm stuck here, any hint?

Answer

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.