A sequence of $a_n, b_n$ is defined as follows. $a_1, b_1 >0$, and $$a_{n+1}=a_n+\frac{1}{b_n}$$ $$b_{n+1}=b_n+\frac{1}{a_n}$$ Prove that $a_{50}+b_{50}>20$
$$\begin{array}{l} \frac{a_{n+1}+b_{n+1}}{2} \\ =\frac{\left(a_n+b_n\right)+\left(\frac{1}{a_n}+\frac{1}{b_n}\right)}{2} \geq \frac{a_n+b_n}{\sqrt{a_n b_n}} \end{array}$$ also, $a_{n+1}+b_{n+1} \geq 4$. Let $c_n = a_n+b_n$ then, $c_n \geq 4$. We need to show $c_{50}>20$. How do we proceed?
(The following solution was inspired by this answer which gives an explicit lower bound for the elements of the sequence defined by $a_{n+1} = a_n + 1/a_n$.)
We have $$ \begin{align} (a_2 + b_2)^2 &= \left(a_1 + \frac{1}{a_1} + b_1 + \frac{1}{b_1}\right)^2 \\ &\ge (2+2)^2 = 16 \end{align} $$ and $$ \begin{align} (a_{n+1}+b_{n+1})^2 &= \left( a_n+b_n + \frac{1}{a_n} + \frac{1}{b_n}\right)^2 \\ &> (a_n+b_n)^2 + 4 + 2 \left(\frac{a_n}{b_n} + \frac{b_n}{a_n}\right) \\ &\ge (a_n+b_n)^2 + 8 \, . \end{align} $$
It follows that for $n > 2$ $$ (a_n + b_n)^2 > (a_2 + b_2)^2 + 8 (n-2) \ge 8n $$ i.e. $$ a_n + b_n > \sqrt{8 n} \, . $$
In particular, $a_{50} + b_{50} > 20$.
Once $a_1$ and $b_1$ are positives, the given sequences are increasing strictly. There's a symmetry between $a_{n}$ and $b_{n}$ in this question.
Case 1:
Let $a_1 \le 1/20$ and $b_1$ be any value, then
$$b_{2} = b_1 + \dfrac{1}{a_1} \ge b_1+ \dfrac{1}{\left(\frac{1}{20}\right)} = b_1 + 20 > 20$$
which leads to $a_2 + b_2 > 20$. As the values increase strictly, then $a_{50}+b_{50} > 20$
Case 2:
Let $a_1 \ge 20$, and $b_1$ be any value, which leads to $$a_1+b_1 > 20$$
As the values increase strictly, then $a_{50}+b_{50} > 20$
Case 3: Other cases with $\frac{1}{20} < a_{1} < 20$
Let $a_1 = b_1 = 1$, then
$$\begin{cases}a_2 = 2 \\ b_2 = 2 \end{cases} \ \ \ \ \ \ \ \ \begin{cases}a_2 = 2.5 \\ b_2 = 2.5 \end{cases} \ \ \ \ \ \ \ \cdots \ \ \ \ \ \ \ \ \begin{cases}a_{50} \approx 10.8 \\ b_{50} \approx 10.8 \end{cases}$$
Then we see that $a_{50}+b_{50} > 20$.
Now, we only need to show that $f_{50}(a_1, b_1) = a_{50}+b_{50}$ gets its minimum at point $(1, 1)$. Proving that, it finish your question.
Doing for $f_{2}$ we get a clue of doing for $f_{50}$:
$$f_2 = a_2 + b_2 = a_1 + b_1 + \dfrac{1}{a_1} + \dfrac{1}{b_1}$$
Finding the minimum of $f_2$ is finding the minimum of $(a_1 + \frac{1}{a_1})$ and $(b_1 + \frac{1}{b_1})$:
$$a_1 + \dfrac{1}{a_1} \ge 2\sqrt{a_1 \cdot \dfrac{1}{a_1} } = 2$$ $$b_1 + \dfrac{1}{b_1} \ge 2\sqrt{b_1 \cdot \dfrac{1}{b_1} } = 2$$
EDIT:
To show that $f_{k}$ gets it's minimum when $a_1 = b_1 = 1$:
$$\begin{align*}f_{k+1}(a_1, b_1) & = a_{k+1}+b_{k+1} = a_{k} + \dfrac{1}{a_{k}} + b_{k} + \dfrac{1}{b_{k}}\\ g_{k}(a_1, \ b_1) & = a_{k}+\dfrac{1}{a_{k}} \\ h_{k} (a_1, \ b_1) & = b_k + \dfrac{1}{b_k}\end{align*}$$
Minimizing $f_{k+1}$ means minimize the independent expressions $g_{k}$ and $h_k$
If $k=1$, the minimium arrives only when $a_{1} = b_{1} = 1$ (due to AM–GM inequality)
For $k\ne 1$, it's not possible to have $a_{k} = b_{k} = 1$.
As $a_{k}$ (and $b_k$) is a strictly increasing sequence, the minimum of $g_{k}(a_1, \ b_1)$ (resp. $h_k$) happens when we find the pair $(a_1, \ b_1)$ such the sequence $a_k$ (resp. $b_k$) keeps as near as possible of $1$.
In other words, it's about selecting the pair $(a_1, \ b_1)$ such the sequence $a_{k}$ (resp. $b_k$) is the smallest increasing sequence.
