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This inequality arose in this question Prove that : $|f(b)-f(a)|\geqslant (b-a) \sqrt{f'(a) f'(b)}$ with $(a,b) \in \mathbb{R}^{2}$ :

$$\forall x>0, \frac {x-1}{\ln(x)} \geq \sqrt{x} $$

Can anybody find a way to prove it without calculus? Like using only inequalities like AM-GM, Jensen or Cauchy – Schwarz?

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**CLARIFICATION**

I want a proof that does not involve setting $g(x)=\frac {x-1}{\ln(x)}- \sqrt{x}$, studying the derivative of $g$ and proving it is $\geq 0$ with this method.

The upvoted answers comply with this criterion.

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For $x>1$, i.e. $\log x>0$, we can write $x=e^z$, and $z>0$. The inequality becomes

$$

z \leq e^{\frac{z}{2}}-e^{-\frac{z}{2}}.

$$

This is trivially true, for example by using the **definition** $$e^u = \sum_{k=0}^\infty \frac{u^k}{k!}.$$

The case $0<x<1$ can be treated in a similar fashion.

I *know* that this proof is not calculus-free, but it stems from the definition of the exponential function. This makes me think that it is not a convexity inequality like all those you refer to in your question. I might be wrong, of course.

Define $t:=\log x$. We have to show that for each $t\neq 0$,

$$\frac{e^t-1}t\geqslant e^{t/2},$$

and, after having multiplied on both sides by $e^{-t/2}$, we are reduced to show that for each positive $t$,

$$e^t-e^{-t}\geqslant 2t.$$

This inequality is easier to handle. We indeed recognize classical hyperbolic sine.

If we know the power series expansion of exponential function, namely, $\exp(t)=\sum_{j=0}^{+\infty}\frac{t^j}{j!}$, the result follows because $e^t-e^{-t}=2\sum_{k=0}^{+\infty}\frac{t^{2k+1}}{(2k+1)!}\geqslant 2\frac{t^{2\cdot 0+1}}{(2\cdot 0+1)!}=2t$.

Letting $x=e^u$, the given inequality is equivalent to

$$1\le {e^u-e^{-u}\over2u}\qquad\text{for }-\infty\lt u\lt\infty$$

(with $u\not=0$ understood, in line with the tacit assumption $x\not=1$). Since the right hand side is invariant under $u\to-u$, it suffices to show

$$u\le{1\over2}(e^u-e^{-u})\qquad\text{for }0\lt u$$

Now at some point we have to use a definition of the exponential function. Let’s use

$$e^u=\lim_{n\to\infty}\left(1+{u\over n}\right)^n=\lim_{n\to\infty}\left(1+u+{n\choose2}\left({u\over n}\right)^2+\cdots+{n\choose n}\left({u\over n}\right)^n\right)$$

Now clearly, when $0\lt u$,

$$u\lt u+{n\choose3}\left({u\over n}\right)^3+{n\choose 5}\left({u\over n}\right)^{5}+\cdots={1\over2}\left(\left(1+{u\over n}\right)^n-\left(1-{u\over n}\right)\right)^n$$

for any $n$, since all the extra terms are positive. (The sum with the binomial coefficients terminates at the last odd number less than or equal to $n$.) Hence $u$ is less than or equal to the limit of the right hand side, which, by definition, is equal to ${1\over2}(e^u-e^{-u})$.

Use $$\log(x)=\int_1^x\frac{dt}{t}.$$ For any $x>0$ the inequality $$\frac{1}{t}\leq 1-\frac{(t-1)}{x}$$ holds for all $t$ between $1$ and $x$ inclusive. Therefore if $x\geq 1$ then $$0\leq\log(x)\leq\frac{(x^{-1}+1)(x-1)}{2}=\frac{x-x^{-1}}{2}$$ and if $0<x\leq 1$ then $$0\geq \log(x)\geq\frac{x-x^{-1}}{2}.$$ The desired result for positive $x\neq 1$ now follows from these inequalities and the fact that $\log(x)=2\log\sqrt{x}$.

A simpler proof is to use derivate.If x > 1 ==> equivalent inequality: x^(1/2) – x^(-1/2) >= 2*ln(x^(1/2)). Let t = x^(1/2) ==>we need to prove: f(t) =t – 1/t – 2*lnt >= 0 for t > 1. f'(t) = 1 + 1/t^2 – 2/t = (t – 1)^2/2t > 0. So f(t) > f(1) = 0. If 0 < x < 1. Let t = 1/x ==> we need to prove:

(1/t – 1)/ln(1/t) >= 1/(t^1/2) <==> (1 – 1/t)/lnt >= 1/(t^1/2) <==> (t – 1)/lnt >= t^(1/2) with t > 1,and we are back to case 1 that has been proved.

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