# I Prove ln(x) <= x-1 for positive x

1. Jan 15, 2017

### 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?

Last edited: Jan 15, 2017
2. Jan 15, 2017

### Math_QED

What is inx?

3. Jan 15, 2017

### Dank2

Natural logarithm ;)

4. Jan 15, 2017

### Math_QED

5. Jan 15, 2017

### Dank2

6. Jan 15, 2017

### Dank2

corrected

7. Jan 15, 2017

### Dank2

solved. thanks

8. Jan 15, 2017

### FactChecker

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

9. Jan 15, 2017

### Dank2

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

10. Jan 15, 2017

### 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.

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