It seems like $$\log n \leq \sqrt n \quad \forall n \in \mathbb{N} .$$ I've tried to prove this by induction where I use $$ \log p + \log q \leq \sqrt p \sqrt q $$ when $n=pq$, but this fails for prime numbers. Does anyone know a proof?
How to prove $\log n \leq \sqrt n$ over natural numbers?
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inequality
logarithms
radicals
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0Does it suffice to note that $f(x)=\log x - \sqrt{x}$ satisfies $f'(x)\leq 0$ for $x\geq 4$ and note that the inequality is satisfied for $n=1,2,3,4$? – 2011-09-19
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1What are you "allowed" to use? (Actually, how are you defining $\log$? I assume you mean the natural log there, and a usual definition is as $\log x = \int_1^x \frac{1}{y} \mathrm{d}y$, so some calculus is used.) – 2011-09-19