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.
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Go ahead and test it, but i'll just guaruntee you it's a min, so:
Min value:
x at which it occurs:
Now how do we provide for all ?
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.