I know that the continuum hypothesis is not decidable, i.e. we can not prove it, nor disprove it. The question is, is it theoretically possible to find an explicit set $E\subset \mathbb R$ such that we can not prove neither $\vert \mathbb N\vert =\vert E\vert$, nor $\vert \mathbb R\vert =\vert E...
I suggest to remove theory-of-equations tag. This tag was created in December. The tag creator also provided the tag-excerpt and the tag-wiki which more-or-less follow the Wikipedia article Theory of equations. Looking at the tag-info, it seems that most of the problems described there are cover...
The tag a.m.-g.m.-inequality has been created recently Is this really going to be useful? It seems a bit too specific to me. In any case, it is probably worth discussing this on meta before the tag grows too large. I am not denying that the inequality is useful and well-known. However, it seem...
To prove $x+\frac{1}{x}\geq2$ where $x$ is a positive real number. This is what i try: $$\text{We need to prove } \hspace{1cm} x+\frac{1}{x}-2\geq 0$$ now,$$\frac{x^2-2x+1}{x}=(x-2)+\frac{1}{x}$$ its enough to show that $$\frac{1}{x}\geq(x-2)\hspace{0.5cm} \text{ when } \hspace{0.2cm}0<x\leq 2$$ ...
If $a,b,c,d>0$ and $abcd=1$ prove: $$\sum \frac{1}{(1+a)(1+a^2)}\ge 1$$ Here's my solution: I try to prove it by reverse: $$3\ge\sum\frac{a^3+a^2+a}{(a+1)(a^2+1)}$$Then by AM-GM: $$3\ge\sum\frac{a(1+a+a^2)}{2a(a+1)}$$$$$6\ge\sum\frac{1+a+a^2}{a+1}$$ Then We need to prove: $$2\ge\sum\frac{a^2}{a...
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