Prove 1x+1&lt;ln⁡(x+1x)&lt;0.5(1x+1x+1)

Question

Prove  1x+1&amp;lt;ln⁡(x+1x)&amp;lt;0.5(1x+1x+1)

Hana Larsen · Accepted Answer

Let us recall a property of integrals, so simple one could think nothing good could come out of it, and yet:
In the present case, one may start from the identity

\log (\frac{(x + 1)}{x}) = \int_{x}^{x + 1} u (t) d t with u (t) = \frac{1}{t}

and note that

u (t) \geq \frac{1}{(x + 1)}

for every

t

in the interval

(x, x + 1)

. This should give one inequality.
For the other inequality, one may prove that

v (s) = u (x + \frac{1}{2} + s) + u (x + \frac{1}{2} - s)

is an increasing function of s for s
in

(0, \frac{1}{2})

and note that

\log (\frac{x + 1}{x}) = \int_{0}^{\frac{1}{2}} v (s) d s

.
Can you reach that point and finish the proof from there?

Karli Kaiser · Accepted Answer

For LHS:use this well konw inequality  xx+1≤ln⁡(1+x),x&amp;gt;−1⇒ln⁡(x+1x)=ln⁡(1+1x)≥1x1+1x  =1x+1  For RHS: we only prove  2ln⁡(1+x)&amp;lt;x+xx+1=x2+2xx+1  let  =−x2(x+1)2≤0  so g(x)≤g(0)=0  so ln⁡(1+x)≤12(x+xx+1)  so ln⁡(x+1x)≤0.5(1x+1x+1)

Prove \frac{1}{x+1}<\ln(\frac{x+1}{x})<0.5(\frac{1}{x}+\frac{1}{x+1})

Answered question

Answer & Explanation

New Questions in Algebra II