<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] Without Calculus:
Multiplying both sides of $a+b+\dfrac 1a + \dfrac 1b \ge 3 \sqrt 2$ by $ab$ we get 
$$a^2b+b^2a+a+b \ge 3 \sqrt2 ab$$
which factors as 
$$(a+b)(ab+1) \ge 3 \sqrt2 ab$$ 
and squaring both sides yields 
$$(a+b)^2(ab+1)^2 \ge 18(ab)^2$$ 
But $(a+b)^2 = 1+2ab$, and substituting $x = ab$ get
$$(1+2x)(x+1)^2\ge18x^2$$
Furthermore, we know $x \in\left [0, \dfrac 12 \right ]$ because $a^2+b^2 \ge 2ab$, so $\dfrac 12 \ge ab$.
Thus it remains to show that $f(x) = (1+2x)(x+1)^2 - 18x^2 \ge 0$ on $\left[0, \dfrac 12 \right]$. But expanding $f(x)$ and using the rational root theorem we can factor it as $f(x) = 2(x-1/2)(x^2-6x-1)$. Now $(x- 1/2)$ is non-positive on the required interval, and you can find the roots of $x^2-6x-1$ using the quadratic formula to show that it is negative on that interval. 
Done!
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