Ln(x) < sqrt(x) for 1<x<infinity

[Belgium 12]

Hi,

How can I show or proof:

1) ln(x)<sqrt(x) for 1<x<infinity

2) ln(x)<1/sqrt(x) for 0<x<1

Thank you

 

[mathman]

1) ln(1)=0, sqrt(1)=1. Take derivatives: 1/x < 1/2sqrt(x).

2) ln(x) for 0<x<1 is negative.

 

[Office_Shredder]

1/x < 1/2sqrt(x)

So when x=1...

you have to be a little bit careful. I think a nice way to approach these is to take the function

$$\sqrt{x}-ln{x}$$ and find the global minimum on $$[1, \infty )$$

and it's not hard to discover that the global minimum has a y value larger than zero