# RMO practice problem inequality

Let $$a_n$$ & $$b_n$$ be two sequences such that $$a_0$$ , $$b_0$$ > 0 and $$a_{n+1}$$ = $$a_n$$ + $$\frac{1}{2b_n}$$ & $$b_{n+1}$$ = $$b_n$$ + $$\frac{1}{2a_n}$$ $$\forall$$ n $$\geq$$ 0. Then prove that $$max(a_{2018},b_{2018}) > 44.$$

• Ok I'll keep that in mind – Mayank Mishra Oct 6 '18 at 19:41

Without loss of generality, suppose that $$a_k>b_k$$ for some k. Then $$a_{k+1}=a_k+\frac{1}{2b_k}>a_k+\frac{1}{2a_k}$$. Therefore it suffices to prove that $$c_{2018}>44$$, where $$c_0$$ is arbitrary positive number and $$c_{n+1}=c_n+\frac{1}{2c_k}$$.
Claim. $$c_n\ge\sqrt{n+1}$$ for all $$n>0$$.
We prove the claim by mathematical induction. For $$n=1$$, the claim is clear since by AM-GM, $$c_{1}=c_0+\frac{1}{2c_0}\ge2\sqrt{\frac{c_0}{2c_0}}=\sqrt{2}$$. For induction case, note that the function $$x+\frac{1}{2x}$$ is increasing for $$x\ge\frac{1}{\sqrt{2}}$$, which means $$c_{n+1}=c_n+\frac{1}{2c_k}\ge\sqrt{n}+\frac{1}{2\sqrt{n}}$$. Therefore it is enough to show that $$\sqrt{n+1}\le\sqrt{n}+\frac{1}{2\sqrt{n}}$$, or $$\sqrt{n+1}-\sqrt{n}\le\frac{1}{2\sqrt{n}}$$, or $$\frac{1}{\sqrt{n+1}+\sqrt{n}}\le\frac{1}{2\sqrt{n}}$$, or $$\sqrt{n+1}+\sqrt{n}\ge2\sqrt{n}$$, which is obvious.
By claim, we know that $$c_{2018}\ge\sqrt{2019}$$ and $$\sqrt{2019}>44$$ and we already showed that $$\max{(a_n,b_n)}>c_n$$ for any $$n$$.