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I can't seem to solve this exercise. I want to show that: $\frac1x+\log x-\log(x+1)\ge0$ for $x\ge1$.

I've tried looking for bounds on $\log x-\log(x+1)$, but there's nothing promising I can derive. I'm thinking of multiplying out $x$, yielding: $$1+x\log x-x\log(x+1)\ge0$$ I believe, I can bound $x\log(x)$ by $ex-e$, but it doesn't seem sufficient, because I cannot find a good bound on $x\log(x+1)$.

I can solve it with differential calculus (indeed, I can check it for $x=0$ then differentiate), but I'd like to see if there is a way to solve it without. We've not been introduced to the formal definitions of differentiation etc. yet.

Parcly Taxel
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  • Exponentiating both sides and transposing leads to the inequality $e^{\frac 1x} \geq 1 + \frac 1x$, which is a case of the inequality $e^y \geq 1+y$ for all $y$. This is well known, but I think it involves calculus. – Sarvesh Ravichandran Iyer Feb 17 '18 at 10:26
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    There are many proofs of the fact that $1+y \leq e^y$ on this page : https://math.stackexchange.com/questions/504663/simplest-or-nicest-proof-that-1x-le-ex, and you may pick the one you think uses least calculus. Then taking $y = \frac 1x$ followed by a logarithm (which preserves inequality, and I think this involves calculus as well) gives your result. Personally, this is just an "in the limit" Bernoulli inequality, as many posts will show you here. – Sarvesh Ravichandran Iyer Feb 17 '18 at 10:33
  • This is only possible if you have defined logarithm without using derivatives /integrals. One possible solution is to use definition $\log x=\lim_{n\to\infty} n(x^{1/n} -1)$ and then your problem is an easy consequence of the fact that the sequence $s_n=n(x^{1/n}-1)$ is decreasing. See this answer https://math.stackexchange.com/a/2646981/72031 for more details. Also you should include the definition of logarithm you are using. – Paramanand Singh Feb 18 '18 at 03:31

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$$\frac1x+\log x-\log(x+1)\ge0$$ $$\iff\frac1x\ge-\log x+\log(x+1)=\log\frac{x+1}x=\log\left(1+\frac1x\right)$$ Define $y=\frac1x$, so that $0<y\le1$: $$\iff y\ge\log(1+y)$$ Now define $z=y+1$ so that $y=z-1$ and $1<z\le2$: $$\iff z-1\ge\log z$$ which is an inequality you have.

Parcly Taxel
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  • The problem is that Taylor series requires differential calculus, which we have not formalized yet. We only have the definition and bounds like $\log(x) \leq x-1$ and $\log(x) \geq (x-1)/x$ for $x > 0$. –  Feb 17 '18 at 10:33
  • @SimonTépoz Luckily that wasn't too hard to fix; another variable definition and it is an inequality you do have. – Parcly Taxel Feb 17 '18 at 10:35
  • Thanks a lot! .. –  Feb 17 '18 at 10:44
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    "We only have the definition and bounds like..." These definitely should be mentioned in the question. – Did Feb 17 '18 at 10:52