Proving an inequality without an integral: 1x+1≤ln⁡(1+x)−ln⁡(x)≤1x

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Proving an inequality without an integral:  1x+1≤ln⁡(1+x)−ln⁡(x)≤1x

ol3i4c5s4hr · Accepted Answer

Let  f(x)=ln⁡(x),  , then the Mean Value Theorem (Differentiation) says that there exists some ξ∈(x,x+1)  ln⁡(1+x)−ln⁡(x)=ln⁡(1+x)−ln⁡(x)(1+x)−x=f(1+x)−f(x)(1+x)−x=dfdxξ  As dfdx(x)=1x  and as the function 1x  is monotonic we know that 1x+1≤dfdx(ξ)≤1x

jean2098 · Accepted Answer

If we write  t=1x  the inequalities become  tt+1≤ln⁡(1+t)≤t,t&amp;gt;0  Observe that the three expressions are equal when  t=0,  and then show, by differentiating, that t−ln⁡(1+t)  and ln⁡(1+t)−t1+t are increasing functions.

Vasquez · Accepted Answer

This is trivial using the definition of ln⁡(x) as the area below the hyperbola 1x between 1 and x. ln⁡(x+1)−ln⁡(x) is the hyperbola area between x and x+1, bounded by the rectangles with area 1(x+1) and 1x

Proving an inequality without an integral: \frac{1}{x+1}\leq\ln(1+x)-\ln(x)\leq\frac{1}{x}

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