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For any two positive real numbers $a, b > 0$, we define:

$$H(x, y) = \frac{x+y}{2\sqrt{xy}} - 1$$

(Note: By the arithmetic mean inequality, $\frac{a+b}{2} \ge \sqrt{ab}$, therefore $H(a, b) \ge 0$ always holds, and $H(x, x)$ = 0 only if $x = x$.)

The real question I want to ask: Is the following theorem of any practical use?

Let $f(x) = x \ln x$, then ...


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Important preliminary note (added 2026-08-19T23:58+02:00): The comments below show (and the downvotes probably suggest) that various people have interpreted my question as some combination of ⓐ representing problem-solving as a “humans versus AI” game, ⓑ wishing to either demonstrate human superiority in this matter or on the contrary to help improve AIs, and...


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$\DeclareMathOperator{\SL}{SL} \DeclareMathOperator{\PSL}{PSL} \DeclareMathOperator{\Vol}{Vol}$ Let $W = H^3/\langle g\rangle$, where $g$ is a single loxodromic element of $\PSL(2,\Bbb C)$ with multiplier of modulus different from $1$. Topologically $W$ is a solid torus; it is a complete hyperbolic manifold of infinite volume, with conformal boundary a torus of modulus tau. I...


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There is much arbitraryness in the Rolfsen (and later) tables. Of course anyone would name $7_1$ to be the first knot with $n=7$ crossings, but already my own "natural" ordering attempt (extending Stasiak's work - I don't know if that has been refined later) suggest at least for rational knots that beginning with $n=8$ the naming is unfortunate. Similar problems may occur for...


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Alice and Bob take turns colouring the edges of a complete graph $K_n$ red and blue respectively, with Alice going first. They continue until the graph is fully coloured.

Next, they take turns removing vertices until a triangle $K_3$ remains. Alice wins if it is monochrome red, Bob wins if it is monochrome blue, otherwise it is a draw.

Question:...


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