amparvardi wrote: Let $x_0 = 5$ and $x_{n+1} = x_n + \frac{1}{x_n} \ (n = 0, 1, 2, \ldots )$. Prove that \[45 < x_{1000} < 45. 1.\] $x_{n+1}^2=x_n^2+\frac 1{x_n^2}+2>x_n^2+2$ and so $x_n^2>25+2n$ $\forall n>0$ and so $x_{1000}^2>2025=45^2$ $x_n^2\ge 25+2n$ $\forall n\ge 0$ and $x_{n+1}^2=x_n^2+\frac 1{x_n^2}+2$ $\implies$ $x_{n+1}^2\le x_n^2+2+\frac 1{2n+25}$ And so $x_{1000}^2\le 2025+\sum_{k=0}^{1000}\frac 1{25+2k}$ And we just have to show that $\sum_{k=0}^{1000}\frac 1{25+2k}<45.1^2-45^2=9.01$, which is quite easy

for example

can anyone tell me , this procedure $x_n^2\ge 25+2n$ $\forall n\ge 0$

siddharthanand wrote: can anyone tell me , this procedure $x_n^2\ge 25+2n$ $\forall n\ge 0$ We have $x_{n}^2>x_{n-1}^2+2$ Summing all these inequalities for $n=1\to n$, we get $x_1^2+x_2^2+...+x_n^2>$ $x_0^2+x_1^2+...x_{n-1}^2+2n$ And so $x_n^2>x_0^2+2n=25+2n$ $\forall n>0$

thank you , Patrick Truly remarkable

did what patrick wrote just solved the problem?

TheLittleEinstein wrote: did what patrick wrote just solved the problem? I dont understand your question : what else should we need ?

pco wrote: amparvardi wrote: Let $x_0 = 5$ and $x_{n+1} = x_n + \frac{1}{x_n} \ (n = 0, 1, 2, \ldots )$. Prove that \[45 < x_{1000} < 45. 1.\] $x_{n+1}^2=x_n^2+\frac 1{x_n^2}+2>x_n^2+2$ and so $x_n^2>25+2n$ $\forall n>0$ and so $x_{1000}^2>2025=45^2$ $x_n^2\ge 25+2n$ $\forall n\ge 0$ and $x_{n+1}^2=x_n^2+\frac 1{x_n^2}+2$ $\implies$ $x_{n+1}^2\le x_n^2+2+\frac 1{2n+25}$ And so $x_{1000}^2\le 2025+\sum_{k=0}^{1000}\frac 1{25+2k}$ And we just have to show that $\sum_{k=0}^{1000}\frac 1{25+2k}<45.1^2-45^2=9.01$, which is quite easy

for example

now is correct $x_{n+1}^2=x_n^2+\frac 1{x_n^2}+2>x_n^2+2$ and so $x_n^2>25+2n$ $\forall n>0$ and so $x_{1000}^2>2025=45^2$ $x_n^2\ge 25+2n$ $\forall n\ge 0$ and $x_{n+1}^2=x_n^2+\frac 1{x_n^2}+2$ $\implies$ $x_{n+1}^2\le x_n^2+2+\frac 1{2n+25}$ And so $x_{1000}^{2}\le 2025+\sum\limits_{k=0}^{999}{\frac{1}{25+2k}}<2025+9.1={{45.1}^{2}}$ And we just have to show that $\sum\limits_{k=0}^{999}{\frac{1}{25+2k}}<5.5<{{45.1}^{2}}-{{45}^{2}}=9.01$, which is quite easy $\sum\limits_{k=0}^{999}{\frac{1}{25+2k}}<\int_{-1}^{999}{\frac{dx}{25+2x}}$ $=\frac{\ln (2023)}{2}-\frac{\ln (23)}{2}$ $<\frac{\ln(2^{11})}2<5.5$

Claim $1$: For all integers $n>0,$ $x_n>\sqrt{25+2n}.$ Proof: Clearly, $x_n$ is positive for all nonnegative integers $n.$ Note that for all integers $n\geq 0,$ $x_{n+1}^2=\left(x_n+\dfrac{1}{x_n}\right)^2=x_n^2+\dfrac{1}{x_n^2}+2.$ Therefore, for all integers $n>0,$ $x_n^2=\displaystyle\sum_{i=0}^{n-1}(x_{i+1}^2-x_i^2)+x_0^2=\displaystyle\sum_{i=0}^{n-1} \left(\dfrac{1}{x_i^2}+2\right)+25>2\cdot n+25,$ so $x_n>\sqrt{25+2n},$ as desired. Now, $x_{1000}>\sqrt{25+2\cdot1000}=\sqrt{2025}=45,$ so it suffices to prove that $x_{1000}<45.1.$ Note that $x_{n+1}-x_n=\dfrac{1}{x_n}$ for all integers $n\geq 0,$ so $x_{1000}=\displaystyle\sum_{n=0}^{999} (x_{n+1}-x_n)+x_0=\displaystyle\sum_{n=0}^{999} \dfrac{1}{x_n}+5.$ Claim $2$: For all nonnegative real numbers $n,$ $\sqrt{25+2n}>\dfrac{\sqrt{26+2n}+\sqrt{24+2n}}{2}.$ Proof: It suffices to prove that $2\sqrt{25+2n}>\sqrt{26+2n}+\sqrt{24+2n}$ for all nonnegative reals $n.$ Since $50+4n=2\sqrt{4n^2+100n+625}>2\sqrt{4n^2+100n+624}=2\sqrt{26+2n}\cdot\sqrt{24+2n},$ $(2\sqrt{25+2n})^2=100+8n=(50+4n)+(50+4n)>50+4n+2\sqrt{26+2n}\cdot\sqrt{24+2n}=(\sqrt{26+2n})^2+2\sqrt{26+2n}\cdot\sqrt{24+2n}+(\sqrt{24+2n})^2=(\sqrt{26+2n}+\sqrt{24+2n})^2.$ Since $2\sqrt{25+2n}$ and $\sqrt{26+2n}+\sqrt{24+2n}$ are positive, we have $2\sqrt{25+2n}>\sqrt{26+2n}+\sqrt{24+2n},$ as desired. Now, for all integers $0\leq n \leq 999,$ we have $x_n\geq \sqrt{25+2n}>\dfrac{\sqrt{26+2n}+\sqrt{24+2n}}{2},$ so $\dfrac{1}{x_n}<\dfrac{2}{\sqrt{26+2n}+\sqrt{24+2n}}=\dfrac{2(\sqrt{26+2n}-\sqrt{24+2n})}{(\sqrt{26+2n})^2-(\sqrt{24+2n})^2}=\sqrt{26+2n}-\sqrt{24+2n}.$ As such, $x_{1000}=\displaystyle\sum_{n=0}^{999} (x_{n+1}-x_n)+x_0=\displaystyle\sum_{n=0}^{999} \dfrac{1}{x_n}+5< \displaystyle\sum_{n=0}^{999} (\sqrt{26+2n}-\sqrt{24+2n})+5=\sqrt{2024}-\sqrt{24}+5.$ Note that $\left(45-\dfrac{1}{90}\right)^2=2025-1+\dfrac{1}{8100}=2024+\dfrac{1}{8100}>2024,$ so $\sqrt{2024}<45-\dfrac{1}{90}.$ Additionally, $\left(4.9-\dfrac{1}{98}\right)^2=24.01-\dfrac{9.8}{98}+\dfrac{1}{98^2}=24.01-0.1+\dfrac{1}{98^2}=24-0.09+\dfrac{1}{98^2}<24-0.09+0.05<24,$ so $\sqrt{24}>4.9-\dfrac{1}{98}.$ As such, $x_{1000}<\sqrt{2024}-\sqrt{24}+5<45-\dfrac{1}{90}-4.9+\dfrac{1}{98}+5=45.1+\dfrac{1}{98}-\dfrac{1}{90}<45.1,$ so $45<x_{1000}<45.1,$ as desired. $\Box$