Undergrad Prove ln(x) <= x-1 for positive x

[Dank2]

proof $\ln x$<=x-1 for positive x.
if i show that at the start of the segment, the value in the start of the segment (0, +inf), of inx is smaller than x-1, and the derivative of inx is always smaller than x-1, is that suffice to proof that inx is always smaller than x-1? if not why?

 

[member 587159]

What is inx?

 

[Dank2]

What is inx?

Natural logarithm ;)

 

[member 587159]

Natural logarithm ;)

Don't you mean $\ln$ then?

https://en.wikipedia.org/wiki/Natural_logarithm

 

[Dank2]

Don't you mean $\ln$ then?

https://en.wikipedia.org/wiki/Natural_logarithm

yes

 

[Dank2]

the value in the start of the segment (0, +inf), of inx is smaller than x-1, AND the derivative of inx is always smaller than x-1,

corrected

 

[Dank2]

solved. thanks

 

[FactChecker]

proof $\ln x$<=x-1 for positive x.
if i show that at the start of the segment, the value in the start of the segment (0, +inf), of inx is smaller than x-1, and the derivative of inx is always smaller than x-1, is that suffice to proof that inx is always smaller than x-1? if not why?

But the derivative of ln(x) is 1/x, which gets huge near 0.

 

[Dank2]

But the derivative of ln(x) is 1/x, which gets huge near 0.

that's right, so:
f(x) = x-1 - Inx . f''(x) = 1 - 1/x = 0 ==> x=1. that should be either maximus or minimum
f(1) = 0.
now if we take the second derivative:
f''(x) = 1/x^2
which means the derivative is increasing constantly and f(1) is the minimum point inthe graph, and hence x-1>=Inx

 

[StoneTemplePython]

It looks like you are ok here.

In my view, the slickest and most intuitive approach to this by far is to use Jensen's Inequality-- as the logarithm is (strictly) negative convex. If you need to, you can derive Jensen's Inequality, and the above relation falls out of it... that is, construct a Taylor Polynomial for the logarithm at x = 1, with the remainder term being O(n^2). I'd suggest using the Lagrange form for the remainder term. Then use the fact that the logarithm has a continuously negative second derivative. Graphically, this means ln(x) will always be below the tangent line for said logarithm evaluated at x = 1, with equality only where x = 1.