Proving the Inequality: logx < x^1/2 for x>1

[ha11]

How can prove logx< x^1/2 for x>1

 

[VeeEight]

What have you tried so far? Perhaps induction will work

 

[awkward]

Here's one approach:

Define f(x) = \sqrt{x} - \ln(x).

See if you can show that f(1) &gt; 0 and f&#039;(x) \geq 0 for x \geq 1.

That would do it. Do you see why?

[Edit] Oops, that won't work, because it isn't true that f&#039;(x) \geq 0 for x \geq 1!

So try this instead-- find the minimum value of f by solving f&#039;(x) = 0. If the minimum is positive (and it is), then you are done.

 

[ha11]

thanks