<s>[INST] 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? 
 [/INST] 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.
</s>