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Jun
13
comment Which Inequality? $\sum_{i=1}^{n}\frac{1}{x_i}\geqslant1$
Indeed, just look at the arithmetic and geometric means of $\dfrac1{x_i}$ (+1)
Jun
13
revised $f^2+(1+f')^2\leq 1 \implies f=0$
added 185 characters in body
Jun
13
comment $f^2+(1+f')^2\leq 1 \implies f=0$
@JackD'Aurizio: I have edited my answer to account for the case that $1/f$ may blow up inside an interval.
Jun
13
comment Must an ideal generated by an irreducible element be a maximal ideal?
I think what freebird is saying is $\langle2\rangle\subsetneq\langle2,x\rangle\subsetneq\mathbb{Z}[x]$
Jun
13
comment Must an ideal generated by an irreducible element be a maximal ideal?
@DietrichBurde: that question deals with PIDs. This is simply a ring.
Jun
13
comment $f^2+(1+f')^2\leq 1 \implies f=0$
@JackD'Aurizio: $1/f$ is $C^1$ except where it goes to $\infty$ ($f=0$). From any non-zero value of $1/f$, the differential inequality forces $1/f$ to zero in a finite time.
Jun
13
comment how many questions did the answer in each round?
This is useful, especially as an approach to handle more complicated problems. (+1)
Jun
13
revised $f^2+(1+f')^2\leq 1 \implies f=0$
note that $(2)$ is true where $f\ne0$
Jun
13
comment $f^2+(1+f')^2\leq 1 \implies f=0$
@JackD'Aurizio: All that matters is that there is a zero of $1/f$ in the interval. If $f$ has a zero, $1/f$ simply goes to $\infty$. The argument is simply that $1/f$ must pass through $0$ and therefore $f$ must go to $\infty$ in a finite interval, which cannot happen since $f\in C^1(\mathbb{R})$.
Jun
13
awarded  Nice Answer
Jun
13
comment $f^2+(1+f')^2\leq 1 \implies f=0$
@AlexanderVigodner: $f=0$ satisfies $(1)$. However, $(2)$ obviously holds wherever $f(x)\ne0$.
Jun
13
revised $f^2+(1+f')^2\leq 1 \implies f=0$
change the interval so that the function passes through $0$
Jun
13
revised $f^2+(1+f')^2\leq 1 \implies f=0$
include the erroneously deleted conclusion
Jun
13
revised $f^2+(1+f')^2\leq 1 \implies f=0$
illustrate where the derivative is used
Jun
13
revised $f^2+(1+f')^2\leq 1 \implies f=0$
add more explanation
Jun
13
comment $f^2+(1+f')^2\leq 1 \implies f=0$
@AlexSchiff: Any function with a slope at least $\color{#C00000}{1/2}$ (e.g. $1/f$) must pass through the $x$-axis at a finite point.
Jun
13
answered $f^2+(1+f')^2\leq 1 \implies f=0$
Jun
12
comment Prove by induction that $(n+1)^2 + (n+2)^2 + … + (2n)^2 = \frac{n(2n+1)(7n+1)}{6}$
wouldn't the $k+1$ case start with $(k+2)^2$? If the equation you state is true, then the general formula is false.
Jun
12
revised Value of the given limit
clarify argument of summation
Jun
12
comment Some questions on Nilpotent matrix
@Debashish: now that this site has a blog, that might be a better place to post expository work.