Prove $ \min \left(a+b+\frac1a+\frac1b \right) = 3\sqrt{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? 

Solution 1
By $H_n \leq G_n \leq A_n \leq Q_n$,
$$a+b+\frac{1}{a}+\frac{1}{b} \ge a+b+\frac{4}{a+b}=(a+b+\frac{2}{a+b})+\frac{2}{a+b} \geq 2\sqrt{2}+\sqrt{\frac{2}{a^2+b^2}} =3\sqrt{2}.$$
Solution 2
$$a+b+\frac{1}{a}+\frac{1}{b} \geq 2(\sqrt{ab}+\frac{1}{2\sqrt{ab}})+\frac{1}{\sqrt{ab}}\geq 2\sqrt{2}+ \sqrt{2}=3\sqrt{2}.$$
