Question

How can I prove that sqrt(x) > ln x when x > 0?

I can quickly see this graphically and intuitively, but how do I do a proof?

Answer 1

How about an argument like this:

the first derivatives of √x and lnx are
1/(2√x) and 1/x respectively.
Neither curve has any max/min points or points of inflection, so they are both increasing all the time
y = √x is clearly greater than y = lnx for 0 < x ≤ 1
(√(a decimal) is positive, ln(decimal less than 1) is negative)

since √x starts off by being above lnx and since both increase , √x stays above lnx , thus √x > lnx

you might want to include the second derivative test
the second derivatives are -1/(4x^(3/2)) and -1/x^2 respectively.

they are both slowing up, but continue to increase, but since x^2 > x^(3/2) for all x > 1 the lnx function is slowing up faster than the square roots function.
Thus the difference between the function actually gets larger and clearly √x > ln x

Answer 2

That works. Thanks so much!