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? 

Calling $a = \cos u, b = \sin u\;\;$ we have
$$
\left(\cos u + \frac{1}{\cos u}\right)+\left(\sin u + \frac{1}{\sin u}\right)\ge 2\sqrt{\left(\cos u + \frac{1}{\cos u}\right)\left(\sin u + \frac{1}{\sin u}\right)} = 2\sqrt{\frac{(\cos^2 u+1)(\sin^2 u + 1)}{\sin u\cos u}}
$$
Now examining 
$$
f(u) = \frac{(\cos^2 u+1)}{\cos u}\frac{(\sin^2 u + 1)}{\sin u}
$$
by symmetry considerations, the feasible minimum is at $u = u_0 = \frac{\pi}{4}\;\;$ (because $\sin u_0 = \cos u_0\;$)
giving the value 
$$
f(\frac{\pi}{4}) = \frac 92
$$
then following we have
$$
2\sqrt{\frac{(\cos^2 u+1)(\sin^2 u + 1)}{\sin u\cos u}}\ge 2\sqrt{\frac 92} = 3\sqrt2
$$
