+0

# Can someone help me understand what I'm doing wrong in an inequality?

0
51
4
+50

Please, can someone just understand me what I'm doing wrong in the following attempt, without proposing other solutions? Thanks!

Problem: suppose $0 My attempt is the following: since$0

Since $6(1+a_n)$ is positive too, multiplying by $6(1+a_n)$ in $\frac{1}{9} < \frac{1}{7+a_n} < \frac{1}{7}$ leads to $\frac{2}{3}(1+a_n) < 6\frac{1+a_n}{7+a_n}<\frac{6}{7}(1+a_n)$.

But as said before, it is $1<1+a_n<3$, so $\frac{2}{3} <\frac{2}{3}(1+a_n) < 6\frac{1+a_n}{7+a_n}<\frac{6}{7}(1+a_n) < \frac{18}{7}$; of course I am wrong, because the function $f(x)=6\frac{1+x}{7+x}$ has indeed maximum $2$ and minimum $\frac{6}{7}$.

I don't understand where I am making mistake, probably I didn't flip correctly an inequality, but I can't see where. Thank you!

Jun 23, 2021
edited by Hitago  Jun 23, 2021
edited by Hitago  Jun 23, 2021
edited by Hitago  Jun 23, 2021

#1
+50
0

I don't know why but I see the question incomplete, I don't know if it is just a problem of mine. I will attach a pic below:

Jun 23, 2021
edited by Hitago  Jun 23, 2021
edited by Hitago  Jun 23, 2021
edited by Hitago  Jun 23, 2021
edited by Hitago  Jun 23, 2021
edited by Hitago  Jun 23, 2021
#2
+121000
+1

If I understand this...let   an  = x

1/9  <    1  /( 7 + x)  <   1/7

1/9  <   1 /( 7 + x)                                     1 / (7 + x)  <   1/7

7  + x    <    9                                                7  <   7 +  x

x  <  2                                                               x  >   0

0  <  x <   2

If we want an integer soliution   x  =  an  = 1

Jun 23, 2021
#3
+50
+1

@CPhill: Thank you, unfortunately the message was incomplete, sorry for the inconvenience. I've attached a picture of the complete message, would you like to answer to it? Thanks and sorry again.

Hitago  Jun 23, 2021
#4
+121000
+1

6/7   <   6 (  1 + x)   / (7 + x)  <   2

6/7  <   6 (1 + x)  / ( 7 + x)              divide through  by  6

1/7  <    (1 + x)   / (7 + x)

1 ( 7 +x)  <   7 (1 + x)

7 +  x  <   7 + 7x

0  <    6x

0  <  x

And

6 (  1 + x)   / (7 + x)  <   2          divide  through  by 6

(1 + x)  /( 7 + x)  <   1/3

3 ( 1 + x)   <   1 (7 + x)

3 +  3x  <  7 + x

2x  <    4

x  <  2

0 <  x  <  2

Which is  what we  assumed

Jun 23, 2021