Since $a_nb_n\geq 0$ for all $n\in \mathbb{Z}^+$ We have $a_{2016} \geq 0$ And since $(a_{n+1}+b_{n+1})^2\geq 4a_{n+1}b_{n+1}$ we get that $\Big( \frac{a_n+b_n}{2}\Big)^2 \geq 4\sqrt{a_nb_n} \rightarrow \Big( \frac{a_n+b_n}{2}\Big)^4 \geq 16a_nb_n$ So $\Big( \frac{a_{n}+b_{n}}{4}\Big)^4 \geq 16\sqrt{a_nb_n} \rightarrow \Big( \frac{a_n+b_n}{4}\Big)^8 \geq 16^2a_nb_n$ So $\Big( \frac{a_n+b_n}{8}\Big)^8 \geq 16^2\sqrt{a_nb_n} \rightarrow \Big( \frac{a_n+b_n}{8}\Big)^{16}\geq 16^4a_nb_n$ We can prove inductively that $\Big( \frac{a_n+b_n}{2^l}\Big)^{2^{l+1}} \geq 16^{2^{l-1}}a_nb_n$ for all $n,l\in \mathbb{Z}^+$ So $\Big( \frac{t+1}{2^l}\Big)^{2^{l+1}} \geq 16^{2^{l-1}}t$ where $t=a_{2016}$ for all $l\in \mathbb{Z}^+$ But $\lim_{l\rightarrow \infty}{\frac{\Big( \frac{c}{2^l}\Big)^{2^{l+1}}}{16^{2^{l-1}}}}=0$ for any constant $c$ So $0\geq a_{2016}$ combine with $a_{2016}\geq 0$, we get $a_{2016}=0$ So for all $n\in \mathbb{Z}^+$, $a_nb_n=0$, we get that $a_{2015}+b_{2015}=2,a_{2014}+b_{2014}=4,...,a_1+b_1=2^{2015}$ So the only possible value of $a_1>0$ is $2^{2015}$ and this obviously possible, for example $a_n=2^{2016-n},b_n=0$ for all $n\in \mathbb{Z}^+$

ThE-dArK-lOrD wrote: Since $a_nb_n\geq 0$ for all $n\in \mathbb{Z}^+$ We have $a_{2016} \geq 0$ And since $(a_{n+1}+b_{n+1})^2\geq 4a_{n+1}b_{n+1}$ we get that $\Big( \frac{a_n+b_n}{2}\Big)^2 \geq 4\sqrt{a_nb_n} \rightarrow \Big( \frac{a_n+b_n}{2}\Big)^4 \geq 16a_nb_n$ So $\Big( \frac{a_{n}+b_{n}}{4}\Big)^4 \geq 16\sqrt{a_nb_n} \rightarrow \Big( \frac{a_n+b_n}{4}\Big)^8 \geq 16^2a_nb_n$ So $\Big( \frac{a_n+b_n}{8}\Big)^8 \geq 16^2\sqrt{a_nb_n} \rightarrow \Big( \frac{a_n+b_n}{8}\Big)^{16}\geq 16^4a_nb_n$ Why? How did you find it?!

Let $U_n=\frac{a_n+b_n}{2},V_n=\sqrt{a_nb_n}$ if for some $n_0:a_{n_0}>0$ and $b_{n_0}>0$ then $a_n>0,b_n>0 \ \forall n>n_0$ so $V_n$ converge to $1$ but $U_n $ converge to 0 which is imposssible so one of the term is 0 then $ U_{2016}=b_{2016} \implies a_1=U_1=2^{2015}$ . RH HAS

Post your solution @YanYau

It is easy to find that $a_n+b_n=\frac{a_1+b_1}{2^{n-1}}$ and $a_nb_n=\sqrt[2^{n-1}]{a_1b_1}$ for all $n\ge 2$. Since $(a_n+b_n)^2\ge 4a_nb_n$, one can find that $\left(\frac{a_1+b_1}{2^{n-1}}\right)^2\ge 4\sqrt[2^{n-1}]{a_1b_1}\Rightarrow (a_1+b_1)^2\ge 2^{2n}\cdot \sqrt[2^{n-1}]{a_1b_1}$. Since this inequality holds for all $n\ge 2$, we must have $a_1b_1=0\Rightarrow b_1=0$. Hence, $0=\sqrt[2^{2015}]{a_1b_1}=a_{2016}\cdot b_{2016}=a_{2016}$. Also, $\frac{a_1}{2^{2015}}=\frac{a_1+b_1}{2^{2015}}=a_{2016}+b_{2016}=1\Rightarrow a_1=2^{2015}$. To show that this value is achievable, one can set that $a_m=2^{2016-m}, b_m=0$ for all odd $m$ and $a_m=0, b_m=2^{2016-m}$ for all even $m$.