Compare A and B

[Albert]

A=$\dfrac {\sqrt {998}+9}{\sqrt{998}+8}$

B=$\dfrac {\sqrt {999}+9}{\sqrt{999}+8}$

Please compare A and B

 

[anemone]

A=$\dfrac {\sqrt {998}+9}{\sqrt{998}+8}$

B=$\dfrac {\sqrt {999}+9}{\sqrt{999}+8}$

My solution:

Spoiler

If we let $f(x)=\dfrac {\sqrt {x}+9}{\sqrt{x}+8}$, differentiate it w.r.t $x$ we get $f'(x)=\dfrac {-1}{2\sqrt{x}(\sqrt{x}+8)^2}$, i.e. $f'(x)<0$ for all real $x$, or more specifically, $f'(x+1)<f'(x)$ and this implies $f(x)>f(x+1)$, hence, we can say that $A=f(998)=\dfrac {\sqrt {998}+9}{\sqrt{998}+8}>B=f(998+1)=\dfrac {\sqrt {999}+9}{\sqrt{999}+8}$.

 

[Albert]

A=$\dfrac {\sqrt {998}+9}{\sqrt{998}+8}$

B=$\dfrac {\sqrt {999}+9}{\sqrt{999}+8}$

Please compare A and B

$ let \,\,x=998$

$A-B$ = $\dfrac{1}{(\sqrt {x}+8)(\sqrt{x+1}+8\big)}\times \big [(\sqrt{x}+9)(\sqrt{x+1}+8\big)- (\sqrt{x+1}+9)(\sqrt{x}+8\big )\big ]$

=$\dfrac{1}{(\sqrt {x}+8\big)(\sqrt{x+1}+8 \big)}\times (\sqrt {x+1} -\sqrt {x}\big)>0$

$\therefore A>B$

 

[soroban]

Hello, Albert!

$A\:=\:\dfrac{\sqrt {998}+9}{\sqrt{998}+8}$

$B\:=\:\dfrac{\sqrt {999}+9}{\sqrt{999}+8}$

[tex]\text{Compare }A\text{ and }B.[/tex]

Let [tex]x = 998[/tex]

. . . . . . . . . . . . . . . . [tex]A \:\gtrless\:B [/tex]

. . . . . . . . . . . [tex]\frac{\sqrt{x}+9}{\sqrt{x}+8} \;\gtrless\;\frac{\sqrt{x+1}+9}{\sqrt{x+1}+8} [/tex]

[tex]\left(\sqrt{x}+9\right)\left(\sqrt{x+1}+8\right) \;\gtrless\;(\sqrt{x}+8)(\sqrt{x+1}+9) [/tex]

[tex]\sqrt{x(x+1)} + 8\sqrt{x} + 9\sqrt{x+1} + 72[/tex]
. . . . . . . . . . . . [tex]\gtrless\:\sqrt{x(x+1)} + 9\sqrt{x} + 8\sqrt{x+1} + 72[/tex]

. . . . . [tex]8\sqrt{x} + 9\sqrt{x+1} \:\gtrless\:9\sqrt{x} + 8\sqrt{x+1}[/tex]

. . . . . . . . . . . .[tex]\sqrt{x+1} \;{\color{red}>}\; \sqrt{x}[/tex]

[tex]\text{Therefore: }\:A \:>\:B[/tex]