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If $p=1$, $N=1$ and $a_1=(1/2,2/3,3/4,4/5,\ldots)$, the infimum equals 1 and is not achieved on a finitely supported vector (moreover, it is not achieved at all). However if $0<p<1$ and the minimize …

Integrate by parts: \begin{align} \int_x^{x+1}\sin(e^t)dt & =\int_x^{x+1}e^{-t}d(-\cos(e^t)) \\ & =e^{-x}\cos e^x-e^{-x-1}\cos e^{x+1}-\int_x^{x+1}e^{-t}\cos e^{t}dt\\ & =e^{-x}\cos e^x-e^{-x-1}\cos e …

Let 0 be the centre of your circle of radius $r$. If $a=2$, we expand the brackets and write $\sum_{i,j} (p_i-p_j)^2=2n\sum p_i^2-2(\sum p_i)^2\leqslant 2nr^2$, with equality if and only if all points …

Let's prove that $$ \arg \frac{1-zf(s-u)}{1-zf(s+u)}< \pi/2- \arg(1-z\bar{z}f(2u)). $$ Then summing this up with an analogous inequality $$ \arg \frac{1-zf(t+u)}{1-zf(t-u)}< \pi/2- \arg(1-z\bar{z}f(- …

Denote $x^2=a^3,y^2=b^3,z^2=c^3$. By AM-GM we have $1+2xyz=1+(abc)^{3/2}+(abc)^{3/2}\geqslant 3\sqrt[3]{1\cdot (abc)^{3/2}\cdot (abc)^{3/2}}=3abc$, so LHS is not less then $$a^3+b^3+c^3+3abc\geqslant …

Put $x_{ij}=x_{ji}$ and consider the symmetriс matrix $A=(x_{ij})$ with zeros on diagonal. Then $\operatorname{tr} A=0$, $\operatorname{tr} A^2=2\sum_{i<j} x_{ij}^2=2$ is fixed, denote it $2=30 \alpha …