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? 
 A: 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.
A: 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.
A: 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!
A: 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 $f(a = 1/\sqrt{2}) = 3 \sqrt{2}$.
A: Calling $a = \cos u, b = \sin u\;\;$ we have
$$
\left(\cos u + \frac{1}{\cos u}\right)+\left(\sin u + \frac{1}{\sin u}\right)\ge 2\sqrt{\left(\cos u + \frac{1}{\cos u}\right)\left(\sin u + \frac{1}{\sin u}\right)} = 2\sqrt{\frac{(\cos^2 u+1)(\sin^2 u + 1)}{\sin u\cos u}}
$$
Now examining 
$$
f(u) = \frac{(\cos^2 u+1)}{\cos u}\frac{(\sin^2 u + 1)}{\sin u}
$$
by symmetry considerations, the feasible minimum is at $u = u_0 = \frac{\pi}{4}\;\;$ (because $\sin u_0 = \cos u_0\;$)
giving the value 
$$
f(\frac{\pi}{4}) = \frac 92
$$
then following we have
$$
2\sqrt{\frac{(\cos^2 u+1)(\sin^2 u + 1)}{\sin u\cos u}}\ge 2\sqrt{\frac 92} = 3\sqrt2
$$
A: We have: $\left(a+b+\dfrac{1}{a}+\dfrac{1}{b}\right)^2=(a+b)^2+2(a+b)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2=1+2ab+4+2\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\dfrac{1+2ab}{(ab)^2}= 5+2ab+\dfrac{2}{ab}+\dfrac{1}{(ab)^2}+\dfrac{2}{ab}= 5+2ab+\dfrac{4}{ab}+\dfrac{1}{(ab)^2}= 1+\left(4+\dfrac{1}{(ab)^2}\right)+ \left(2ab+\dfrac{4}{ab}\right)\ge 2\sqrt{4\cdot\dfrac{1}{(ab)^2}}+(a+b)^2+\dfrac{4}{\dfrac{(a+b)^2}{4}}=\dfrac{4}{ab}+(a+b)^2+\dfrac{16}{(a+b)^2}\ge \dfrac{4}{\dfrac{a^2+b^2}{2}}+\dfrac{((a+b)^2-8)((a+b)^2-2)}{(a+b)^2}+10= 8+10+\dfrac{(a+b)^2-8)((a+b)^2-2)}{(a+b)^2}\ge 18$ because $0<(a+b)^2\le 2(a^2+b^2) = 2\implies a+b+\dfrac{1}{a}+\dfrac{1}{b}\ge \sqrt{18}=3\sqrt{2}$ , with $=$ occurs at $a = b = \dfrac{1}{\sqrt{2}}$
A: 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, $a=b=\dfrac{\sqrt{2}}{2}.$
A: Notice
$$\begin{align}
  a + b + \frac1a + \frac1b
= &\; \left(a + \frac{1}{2a} + \frac{1}{2a}\right)
+ \left(b + \frac{1}{2b} + \frac{1}{2b}\right)\\
\\
\color{blue}{\rm AM \ge \rm GM \rightarrow\quad} 
\ge &\;  3\left[\left(\frac{1}{4a}\right)^{1/3} + \left(\frac{1}{4b}\right)^{1/3}\right]\\
\color{blue}{\rm AM \ge \rm GM \rightarrow\quad} 
\ge &\; 6 \left(\frac{1}{16ab}\right)^{1/6}\\
\color{blue}{a^2 + b^2 \ge 2 ab \rightarrow\quad}
\ge &\; 6 \left(\frac{1}{8(a^2+b^2)}\right)^{1/6}\\
= &\; 6 \left(\frac18\right)^{1/6} = \frac{6}{\sqrt{2}} = 3\sqrt{2}
\end{align}
$$
Since the value $3\sqrt{2}$ is achieved at $a = b = \frac{1}{\sqrt{2}}$, we have
$$\min \left\{ a + b + \frac1a + \frac1b : a, b > 0, a^2+b^2 = 1 \right\} = 3\sqrt{2}$$
Notes
About the question how do I come up with this. I basically know the minimum is
achieved at $a = b = \frac{1}{\sqrt{2}}$. Since the bound $3\sqrt{2}$ on RHS of is optimal, if we ever want to prove the inequality, we need to use something that is tight when $a = b$. If we want to use AM $\ge$ GM, we need to arrange the pieces so that all terms are equal. That's why I split $\frac1a$ into $\frac{1}{2a} + \frac{1}{2a}$ and $\frac1b$ into $\frac{1}{2b} + \frac{1}{2b}$ and see what happens. It just turns out that works.
A: Solution 1
By $H_n \leq G_n \leq A_n \leq Q_n$,
$$a+b+\frac{1}{a}+\frac{1}{b} \ge a+b+\frac{4}{a+b}=(a+b+\frac{2}{a+b})+\frac{2}{a+b} \geq 2\sqrt{2}+\sqrt{\frac{2}{a^2+b^2}} =3\sqrt{2}.$$
Solution 2
$$a+b+\frac{1}{a}+\frac{1}{b} \geq 2(\sqrt{ab}+\frac{1}{2\sqrt{ab}})+\frac{1}{\sqrt{ab}}\geq 2\sqrt{2}+ \sqrt{2}=3\sqrt{2}.$$
