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? 

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.
