use Calculus to find the minimum value of Sqrt(x)-lnx and the value of x for which it occurs.Hence show Sqrt(x)>lnx for all x>0. Thanx..
Your minimum value is going to occur at all x such that f'(x)=0 when f'(x - a) < 0 and f'(x + a) > 0.
$\displaystyle f(x) = \sqrt x - lnx$
$\displaystyle f'(x) = \frac{1}{2\sqrt x} - \frac{1}{x} = \frac{\sqrt x - 2}{x}$
$\displaystyle \sqrt x - 2 = 0$ @ $\displaystyle x = 4$
Go ahead and test it, but i'll just guaruntee you it's a min, so:
Min value: $\displaystyle \sqrt 4 - ln4$
x at which it occurs: $\displaystyle x = 4$
Now how do we provide $\displaystyle \sqrt x > lnx$ for all $\displaystyle x > 0$?
Well, our minimum is giving us a point when x^(1/2) and lnx are closest (being the lowest y on the graph, we must have the smallest difference between the functions). So if we test x=4 for both functions individually and find that one is greater than the other, we've proved this.