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Prove that $$\ln x\leq\frac{x^{x+1/x}-1}{2}$$ is true for every positive real number, without calculus/derivative . (i.e. using some inequalities)

My progress. For $x\geq 1$ using $x+1/x\geq 2$ we obtain $$\frac{x^{x+1/x}-1}{2}\geq\frac{x^{2}-1}{2}$$ So it suffices to prove $$\ln x\leq\frac {x^2-1}{2}$$ or $$x^2\geq 2\ln x+1$$ The inequality seems easier now. But, I am stuck here.

I don't have any approach for $0<x<1$.

user1210867
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5 Answers5

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Update: Thank @lone student for pointing out my mistake. I give a new proof.

Supplement to Alex Ravsky's answer.

We use the well-known inequality $$\mathrm{e}^u \ge 1 + u, \quad \forall u \in \mathbb{R} \tag{1}$$ and the Bernoulli inequality $$(1 + u)^r \ge 1 + ur, \quad \forall u > -1,\ r \ge 1. \tag{2}$$

Using (1), we have $\ln x \le x - 1$. It suffices to prove that $$x^{x + 1/x} + 1 - 2x \ge 0. \tag{3}$$

Alex Ravsky has proved the case that $0 < x \le 1/2$ and $x\ge 1$.

It remains to prove the case that $1/2 < x < 1$.

Using (2) and $x + 1/x - 1 \ge 1$, we have $$x^{x + 1/x - 1} \ge 1 + (x - 1)(x + 1/x - 1). \tag{4}$$

From (3) and (4), it suffices to prove that $$x\cdot [1 + (x - 1)(x + 1/x - 1)] + 1 - 2x \ge 0$$ or $$x(x - 1)^2 \ge 0$$ which is true.

We are done.

River Li
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For $x>0$ we have $\ln x\le x-1 $ and $x^{{x^2+1\over x}}\ge x^2$ for $x\ge 1.$ So it suffices to show that $2x-2\le x^2-1.$ The latter reduces to $(x-1)^2\ge 0.$ The justification of the inequality $\ln x\le x-1$ for $x> 0$ requires calculus methods. see

Ryszard Szwarc
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This answer is partial.

Consider the following graphs.

enter image description here

They suggest to split the required inequality into the inequalities $\ln x\le x-1$ and $x-1\le \frac{x^{x+1/x}-1}2$. Then first of them follows from the inequality $1+t\le e^t$, which holds for each real $t$. The second transforms into the inequality $2x-1\le x^{x+1/x}$. The latter holds when $0<x\le \tfrac{1}{2}$, because then the left-hand side nonpositive, whereas the right-hand side is nonnegative. The inequality also holds when $x\ge 1$, because then $x^{x+1/x}\ge x^2\ge 2x-1$.

Alex Ravsky
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    Moreover, $y=2x-1$ is the common tangent line at $x=1$ to both $y=2\ln x-1$ and $y=x^{x+1/x}$. It is easy to show that $2\ln x-1$ is concave down on $(0,\infty)$, so the difficult part is to show (by hand) that $y=x^{x+1/x}$ is concave up on $(0,\infty)$. Mathematica shows that right away, since no roots of the second derivative of $x^{x+1/x}$ are real, and thus the convexity does not change. A bit of calculus also shows that the second derivative of $x^{x+1/x}$ at $x=1$ is $1>0$, so $x^{x+1/x}$ is concave up on $(0,\infty)$. – Alexander Burstein Aug 25 '23 at 01:20
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$$ \frac{x^{x+\frac{1}{x}}-1}2= \frac{e^{ln(x)(x+1/x)}-1}2 \ge \frac{1+(ln(x)(x+\frac{1}{x})-1}2 = 0.5ln(x)(x+\frac{1}{x}) \ge ln(x)$$

I used first order taylor of $e^x$.

And Am-Gm inequality.

only for $$x \ge 1$$

eyal20
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    I think a little bit more work needs to be done justifying the first inequality because $\ln x$ could be negative and then things are a bit suspicious. – TheBestMagician Aug 27 '23 at 14:27
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For $\frac1{\sqrt{e}}<x<1:$

The inequality can be written as $$\ln(1+2\ln x)\leq(x+\frac1x)\ln x.$$ On the left hand side, we use Maclaurin series approximation and then cancel $\ln x$ from both sides. Now, it is enough to prove that $$2-2\ln x\geq x+\frac1x$$ on $I=(\frac1{\sqrt{e}}, 1)$. Or $$2+2t>e^t+e^{-t}$$ on $(0,\frac12)$ which is true.

Bob Dobbs
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