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From the graph it looks like $(1+1/x)^x$ is concave for $x>0$. But in this post, I can only prove that it is concave for $x\ge 1$. It is of interest to see a proof for $x>0$.

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Considering the second derivative, it is sufficient to prove that

$$(x+1) \log (1 + 1/x) - 1 \le \frac{1}{\sqrt{x}}$$

We will show that,

$$\log (1 + 1/x) \le \frac{1}{\sqrt{x}}$$

This implies, for $0 \lt x \le 1$, that

$$(x+1) \log (1 + 1/x) - 1 \le \frac{x+1}{\sqrt{x}} - 1 = (\sqrt{x} - 1) + \frac{1}{\sqrt{x}} \le \frac{1}{\sqrt{x}}$$

We can easily show that $f(y) = y - \log(1+y^2)$ is increasing (by considering the derivative), and this implies that $f(y) \ge f(0) = 0$ for $y \ge 0$, and thus $$\log (1 + 1/x) \le \frac{1}{\sqrt{x}}$$

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