<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] This is a summary of the arguments and comments above:
$a^2 + b^2 = 1$ and they symmetry argument that $a = b$ shows that $a = b = {1 \over \sqrt{2}}$, and thus $a + b + 1/a + 1/b = 3 \sqrt{2}$.
Calculus confirms it:
$$f(a) = a + 1/a + \sqrt{1 - a^2} + {1 \over \sqrt{1 - a^2}}$$
so
$${df(a) \over da} = -\frac{a}{\sqrt{1-a^2}}+\frac{a}{\left(1-a^2\right)^{3/2}}-\frac{1}{a^2}+1$$
Set this equal to zero and solve for $a$ to find $a = {1 \over \sqrt{2}}$ and the rest follows.
Here's a plot of $f(a)$:

Indeed, the minimum occurs at $a = {1 \over \sqrt{2}}$ and has value $f(a = 1/\sqrt{2}) = 3 \sqrt{2}$.
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