Multiplying the recurrence, we have $a_{n+1}b_{n+1}=1+a_nb_n+ \frac{1}{4a_nb_n} \ge 2$. This implies that $a_nb_n > n+1$ for all $n \in \mathbb{N}$ so $\text{max}\left(a_{2018}, b_{2018}\right) >\sqrt{2019} > 44$.

Let $a_1\geq b_1$ then $a_2 \geq b_2 \to a_n \geq b_n$ $2a_2 \geq a_2+b_2 =a_1+\frac{1}{2a_1}+b_1+\frac{1}{2b_1} \geq 2\sqrt{2} \to a_2 \geq \sqrt{2}$ By induction, let $a_n \geq \sqrt{n}$ then $a_{n+1}=a_n+\frac{1}{2b_n} \geq a_n+\frac{1}{2a_n} \geq \sqrt{n}+\frac{1}{2\sqrt{n}}=\sqrt{n+1+\frac{1}{4n}}>\sqrt{n+1}$ so $a_{n+1} > \sqrt{n+1}$ So $a_{2018}>\sqrt{2018}$

When $n\geq2,$ $$a_nb_n\geq n.$$

Vfire wrote: Multiplying the recurrence, we have $a_{n+1}b_{n+1}=1+a_nb_n+ \frac{1}{4a_nb_n} \ge 2$. This implies that $a_nb_n > n+1$ for all $n \in \mathbb{N}$ so $\text{max}\left(a_{2018}, b_{2018}\right) >\sqrt{2019} > 44$. I guess u make a mistake. $a_nb_n>n+1$ is not necessary true it will $a_nb_n\ge n$.since $a_n,b_n$ are real

ftheftics wrote: Vfire wrote: Multiplying the recurrence, we have $a_{n+1}b_{n+1}=1+a_nb_n+ \frac{1}{4a_nb_n} \ge 2$. This implies that $a_nb_n > n+1$ for all $n \in \mathbb{N}$ so $\text{max}\left(a_{2018}, b_{2018}\right) >\sqrt{2019} > 44$. I guess u make a mistake. $a_nb_n>n+1$ is not necessary true it will $a_nb_n\ge n$.since $a_n,b_n$ are real It's correct. They are positive reals only

Observe that $a_2b_2 = a_1b_1 + \frac{1}{a_1b_1} + 1$ and so $a_2b_2 \ge 2$ by AM-GM. For $n \ge 2$, $a_{n+1}b_{n+1} = a_nb_n +1 + \frac{1}{a_nb_n} > a_nb_n+1$. So, by induction we get that $a_nb_n > n$ for all $n \ge 3$. So since $a_{2018}b_{2018} > 2018$, it follows that max$(a_{2018},b_{2018}) > \sqrt{2018} >44$

Slightly different solution. Let $c_n=a_n+b_n$. Adding the equations gives us $c_{n+1}=c_n+\frac{2c_n}{4a_nb_n}\geq{}c_n+\frac{2}{c_n}$. Squaring, we get that $c_{n+1}^2=c_n^2+4+\frac{4}{c_n^2}>c_n^2+4$, which implies that $c_{2018}>\sqrt{4*2017}>88$, the conclusion follows.

If $a_n\ge b_n$ for some $n$, then: $$a_{n+1}=a_n+\frac1{2b_n}\ge a_n+\frac1{2a_n}\ge b_n+\frac1{2a_n}=b_{n+1}.$$WLOG $a_1\ge b_1$, then $a_n\ge b_n$ for all $n\in\mathbb N$. Now we make use of the inequality $a_{n+1}\ge a_n+\frac1{2a_n}$. We claim that $a_n>\sqrt{n-\frac32}$ for each $n\ge2$. For the base case, note that $a_2\ge a_1+\frac1{2a_1}\ge\sqrt2>\frac{\sqrt2}2$ by AM-GM, so $a_n>\frac{\sqrt2}2$ for all $n\ge2$. Also, for the induction step, we have (after assuming that $a_n>\sqrt{n-\frac32}$): $$a_{n+1}\ge a_n+\frac1{2a_n}>\sqrt{n-\frac32}+\frac1{2\sqrt{n-\frac32}}$$(since $x+\frac1{2x}$ is strictly increasing for $x>\frac{\sqrt2}2$), so it suffices to show that: $$\sqrt{n-\frac32}+\frac1{2\sqrt{n-\frac32}}>\sqrt{n-\frac12}$$for $n\ge2$. This is equivalent to: $$\sqrt n+\frac1{2\sqrt n}>\sqrt{n+1}$$for $n\ge\frac12$, or (after squaring both sides): $$n+1+\frac1{4n}>n+1.$$But this is true from $\frac1{4n}>0$. Induction done. Finally: $$a_{2018}>\sqrt{\frac{4033}2}>\sqrt{\frac{3872}2}=44.$$ The bound can probably be improved to at least $45$.