Just double the other

We can get some loose bounds by truncating the fractions.

$\begin{aligned} A = 1 + \frac{1}{ 1 + \left [ {\frac{ 1}{ 1 + \frac{ 1} { \ddots } } } \right ] } \end{aligned}$

The quantity in the square brackets is clearly smaller than $1$ (we have $1$ divided by something larger than $1$ ); it follows that $A>1+\frac{1}{1+1}=\frac{3}{2}$

Similarly, we have

$\begin{aligned} B = 2 + \frac{1}{ 2 + \left [ {\frac{ 1}{ 2 + \frac{ 1} { \ddots } } } \right ] } \end{aligned}$

Here, we use the fact that the bracketed quantity is greater than zero to get $B<2+\frac{1}{2} = \frac{5}{2}$

Hence $\boxed{2A>B}$

This needs to be under "Basic."

The following are readily observable from inspecting the first line of the continued

fraction for A and the second line of the continued fraction for B:

A > 1.5

Then, 2A > 3

But, B < 2 + 1 = 3

Therefore, 2A is larger.

$A=1+\frac {1}{A}$

Do the same for B and compare.

$A=0.5(1+√5)$ and $B=1+√2$

[This is not a complete solution.]

For those who know how to solve the continued fraction, we can show that $A = \frac{1 + \sqrt{5} } {2}$ and $B = 1 + \sqrt{2}$ .
From here, it directly follows that $2A > B$ .

How can we show this directly by just comparing fractions?
Cryptic Hint: What does $2A$ look like in the above format?

---

(More details in comments) Cryptic hint elaborated on:
If you naively tried to find $2A$ , you will say that it looks like

$2A = 2 + \frac{1}{ \frac{1}{2} + \frac{ 1}{ 2 + \frac{ 1} { \ddots } } }$

However, this isn't a continued fraction, which requires that all of the terms in the fraction are positive integers. Furthermore, this doesn't quickly allow for an easy comparison with $B$ .

As such, this isn't easy to work with. What's the next step in trying to work out this problem?

We can get these equations A=1+ $\frac{1}{A}$ and B=2+ $\frac{1}{B}$ by substituting the infinite fraction as it is a same repeating pattern. $A^2-A-1=0$ and $B^2-2B-1=0$ So the roots are $A=(1+\sqrt 5)/2$ and $B=1+\sqrt 2$ . Hence 2A>B ( A and B are greater than 0 hence we cannot have $A=(1-\sqrt 5)/2$ and $B=1-\sqrt 2$ )

Let $A = 1 + R,$ where $R$ is the continued fraction. Then,

$R = 1/(1+R)$ (the continued fraction equals its subfraction.)

$R + R^2 = 1$

$R^2 + R - 1 = 0$

apply quadratic formula:

$R = \frac{-1 \pm sqrt (5)}{2}$

$R = 0.61803\dots$ (choosing the positive solution)

Let $B = 2 + S,$ where $S$ is the continued fraction. Then,

$S = 1/(2+S)$

$2S + S^2 = 1$

$S^2 + 2S - 1 = 0$

apply quadratic formula:

$S = \frac{-2 \pm sqrt(8)}{2}$

$S = 0.41421...$ (choosing the positive solution)

Then $A = 1.61803\dots,$ $B = 2.41421\dots,$ $2A = 3.2361\dots$ and $2A > B$ .

$A = \frac{1+\sqrt{5}}{2}$ and $B = 1+\sqrt{3}$ . Then $2A = 1+\sqrt{5}$ and $B = 1 + \sqrt{3}$ so $2A > B$ .

Method 1: This question is essentially asking us to prove that $\dfrac AB > \dfrac12$ . I will prove a stronger bound, $\dfrac AB > \dfrac23$ .

Using the same idea as Chris Lewis' solution, we have

$\begin{aligned} A\quad\quad\quad\quad &>& 1 + \dfrac1{1+1} = \dfrac32 \\ A = 1 + \dfrac1A &<& \dfrac53 \\ A = 1 + \dfrac1A &>& \dfrac85 \\ A = 1 + \dfrac1A &<& \dfrac{13}8 \\ A = 1 + \dfrac1A &>& \dfrac{21}{13}. \end{aligned}$

Similarly, $\begin{aligned} B \quad\quad\quad\quad &<& 2 + \dfrac1{2} = \dfrac52 \\ B = 2 + \dfrac1B &> & \dfrac{12}5 \\ B = 2 + \dfrac1B &<& \dfrac{29}{12}. \end{aligned}$

Because $A> \dfrac{21}{13} > \dfrac{29}{18} \Rightarrow 6A > \dfrac{29}3 ,$ and $4B < \dfrac{29}3$ , putting it all together gives $6A > \dfrac{29}3 > 4B \qquad\Rightarrow \qquad 6A > 4B \qquad\Rightarrow \qquad \dfrac AB > \dfrac23.$

Since $\dfrac AB > \dfrac23 > \dfrac12$ . It follows that $2A > B$ . Hence, the answer is $\boxed{2A}$ .

---

Method 2: We want to prove that $2A > B$ , which is equivalent to proving that $10A - 5B > 0 \qquad \Leftrightarrow \qquad (10A - 2) - (5B - 2) > 0 .$

It's obvious that $1 < A < 2$ and $2 < B < 3$ , so $A < B \Rightarrow \frac 1A > \frac 1B \Rightarrow \frac1A - \frac1B > 0$ .

With $10A - 2 = 10 \left( 1 + \frac{1}{ 1 + \frac{ 1}{ 1 + \frac{ 1} { \ddots } } } \right) - 2 = 8 + \frac{10}{ 1 + \frac{ 1}{ 1 + \frac{ 1} { \ddots } } },$ and $5B - 2 = 5 \left( 2 + \frac{1}{ 2 + \frac{ 1}{ 2 + \frac{ 1} { \ddots } } } \right) - 2 = 8 + \frac{5}{ 2 + \frac{ 1}{ 2 + \frac{ 1} { \ddots } } } .$

Putting them together gives $(10A - 2) - (5B - 2) = 5\left( \frac{1}{ 1 + \frac{ 1}{ 1 + \frac{ 1} { \ddots } } } - \frac{1}{ 2 + \frac{ 1}{ 2 + \frac{ 1} { \ddots } } } \right) + 5 \left( 1 + \frac{1}{ 1 + \frac{ 1}{ 1 + \frac{ 1} { \ddots } } } \right) = 5\underbrace{\left( \frac1A - \frac1B\right)}_{> \; 0} + 5A > 0 .$

And we're done!

I just pretended that the fractions weren't infinite so a was 1+1/1 and b was 2+1/2. If you multiplied a by 2 it would be more. I'm not sure if this is how you're supposed to do it though.