Proposed solution of problem 4264

Let (a_{n}) and (b_{n}) be two sequences such that a_{0}, b_{0} > 0 and

a_{n+1} = a_{n} + \frac{1}{2b_{n}} and b_{n+1} = b_{n} + \frac{1}{2a_{n}}

for all n ≥ 0. Prove that

max(a_{2017}, b_{2017}) > 44

Proof The quantity

f(a_{n}, b_{n}) = \frac{a_{n}}{b_{n}}

is invariant respect to the flow of the recurrence. Indeed

f(a_{n+1}, b_{n+1}) = \frac{a_{n} + \frac{1}{2b_{n}}}{b_{n} + \frac{1}{2a_{n}}} = \frac{a_{n}}{b_{n}} = f(a_{n}, b_{n})

This means that

\frac{a_{n}}{b_{n}} = f(a_{n}, b_{n}) = f(a_{0}, b_{0}) = \frac{a_{0}}{b_{0}}

and then

a_{n+1} = a_{n} + \frac{1}{2a_{n}} \frac{a_{0}}{b_{0}} and b_{n+1} = b_{n} + \frac{1}{2b_{n}} \frac{b_{0}}{a_{0}}

Thus

a_{n+1}^{2} + b_{n+1}^{2} = a_{n}^{2} + b_{n}^{2} + \underbrace{(\frac{a_{0}}{b_{0}} + \frac{b_{0}}{a_{0}} )}_{≥2 (AGM)} + \underbrace{(\frac{1}{4a_{n}^{2}} \frac{a_{0}^{2}}{b_{0}^{2}} + \frac{1}{4b_{n}^{2}} \frac{b_{0}^{2}}{a_{0}^{2}} )}_{≥0} ≥ a_{n}^{2} + b_{n}^{2} + 2

a_{n+1}^{2} + b_{n+1}^{2} > 2(n + 1) + a_{0}^{2} + b_{0}^{2} > 2(n + 1)

Moreover

max(a_{2017}, b_{2017}) ≥ \begin{matrix}√ & \overline{a_{2017}^{2} + b_{2017}^{2}} \\ √ & \overline{2} \end{matrix} > \sqrt{2017} > 44.9

and this completes the proof.

Paolo Perfetti