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 $$f(1) > 0$$ and $$f'(x) \geq 0$$ for $$x \geq 1$$.

That would do it. Do you see why?

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

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

 

[ha11]

thanks