<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] The following solution may be the most essential solution...
Denote $$f(x)=\sqrt{x}+\frac{1}{\sqrt{x}}, ~~~x \in (0,1).$$
Since $$f''(x)=\dfrac{3-x}{4x^{5/2}}>0,~~~\forall x \in (0,1),$$ hence $f(x)$ is a  convex function over $(0,1)$. Notice that $a^2, b^2 \in (0,1).$Therefore, $$a+b+\frac{1}{a}+\frac{1}{b}=f(a^2)+f(b^2) \geq 2f\left(\frac{a^2+b^2}{2} \right)=2f\left(\frac{1}{2}\right)=3\sqrt{2},$$with the equality holding iff $a^2=b^2$, namely, $a=b=\dfrac{\sqrt{2}}{2}.$
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