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 $$max(a_{2018},b_{2018}) > 44.$$

Anyone Please help me with this question.. How to approach this?

  • $\begingroup$ Ok I'll keep that in mind $\endgroup$ – 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 $n$.

  • $\begingroup$ Amazing solution. Is it your original solution or you knew it before hand a proof very similar to this $\endgroup$ – Satish Ramanathan Oct 6 '18 at 15:47

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