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

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Discussion Overview

The discussion revolves around proving the inequality ##\ln(x) \leq x - 1## for positive values of x. Participants explore various approaches to establish the validity of this inequality, including derivative analysis and the application of Jensen's Inequality.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that showing the value of ##\ln(x)## at the start of the interval (0, +∞) is smaller than ##x - 1## and that the derivative of ##\ln(x)## is always smaller than ##x - 1## could suffice for the proof.
  • Another participant questions the meaning of "inx," which is clarified to refer to the natural logarithm, ##\ln(x)##.
  • A participant points out that the derivative of ##\ln(x)##, which is ##1/x##, becomes large as x approaches 0, raising concerns about the behavior of the function near this point.
  • Further analysis involves defining a function ##f(x) = x - 1 - \ln(x)## and calculating its second derivative to find critical points, suggesting that ##f(1) = 0## indicates a minimum point.
  • Another participant introduces Jensen's Inequality, arguing that it provides an intuitive approach to the problem, noting the logarithm's negative convexity and suggesting the use of Taylor Polynomials for further exploration.

Areas of Agreement / Disagreement

Participants express differing views on the sufficiency of various approaches to prove the inequality, with no consensus reached on a definitive method or conclusion.

Contextual Notes

Some participants' arguments depend on the behavior of the logarithm near zero and the implications of the second derivative, which remain unresolved. The discussion includes multiple approaches and interpretations without a clear resolution.

Dank2
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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?
 
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What is inx?
 
Math_QED said:
What is inx?
Natural logarithm ;)
 
Dank2 said:
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
 
solved. thanks
 
Dank2 said:
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.
 
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FactChecker said:
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
 
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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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