Ln(x) < sqrt(x) for 1<x<infinity

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SUMMARY

The discussion focuses on proving two inequalities involving the natural logarithm and square root functions: ln(x) < sqrt(x) for 1 < x < ∞ and ln(x) < 1/sqrt(x) for 0 < x < 1. The first inequality is established by taking derivatives, showing that 1/x < 1/(2sqrt(x)). The second inequality is supported by noting that ln(x) is negative for 0 < x < 1, reinforcing that 1/x < 1/(2sqrt(x)). The global minimum of the function sqrt(x) - ln(x) on the interval [1, ∞) is also analyzed, confirming that it remains above zero.

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  • Understanding of calculus, specifically derivatives
  • Familiarity with the properties of logarithmic and square root functions
  • Knowledge of global minima in mathematical functions
  • Basic comprehension of inequalities in real analysis
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  • Learn about derivatives and their applications in proving inequalities
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Belgium 12
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Hi,

How can I show or proof:

1) ln(x)<sqrt(x) for 1<x<infinity

2) ln(x)<1/sqrt(x) for 0<x<1

Thank you
 
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1) ln(1)=0, sqrt(1)=1. Take derivatives: 1/x < 1/2sqrt(x).

2) ln(x) for 0<x<1 is negative.
 


mathman said:
1/x < 1/2sqrt(x)

So when x=1...

you have to be a little bit careful. I think a nice way to approach these is to take the function

\sqrt{x}-ln{x} and find the global minimum on [1, \infty )

and it's not hard to discover that the global minimum has a y value larger than zero
 

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