Hermite Id.

[rsoldo]

If $x$ is a positive number and $n$ is a positive integer, prove using the Hermite's Identity that:

$$\[[nx] > \frac{[x]}{1} + \frac{[2x]}{2} + \frac{[3x]}{3} + ... + \frac{[nx]}{n}.\]$$

[wrest]

rsoldo
Inequality does not hold for $x=1,n=1$, we have $nx=1; [nx]=[1\cdot x]/1$ (assume square brackets mean integer part of a number between them)

[rsoldo]

It's a trivial case when hold equality

[wrest]

rsoldo
Yes, but you’ve used $>$ symbol, not $\ge$
Mistake?

[nnosipov]

rsoldo в сообщении #1408298 писал(а):

If $x$ is a positive number and $n$ is a positive integer, prove using the Hermite's Identity that:

$$\[[nx] > \frac{[x]}{1} + \frac{[2x]}{2} + \frac{[3x]}{3} + ... + \frac{[nx]}{n}.\]$$

In the case $\{x\}<1/n$, this is not true. Can you give a correct version?

[rsoldo]

Ok. This is another case where hold equality. My mistake, I have to include equality.
Otherwise, this is Problem from USAMO 1981.

[rsoldo]

Reminder on Problem if anyone got nice idea ( using the Hermite's Identity, of course ).


If $x$ is a positive real number, and $n$ is a positive integer, prove that

$$\[[ nx]\geq \frac{[ x]}1 + \frac{[ 2x]}2 +\frac{[ 3x]}3 + \cdots + \frac{[ nx]}n\]$$

Спасибо!

[Edward_Tur]

Индукция. Для $n=1$ неравенство очевидно. Индуктивный переход. Предположим, что оно верно для $1,2,...,n$ и докажем его для $n+1$.
Сложим неравенства $[x]\ge[x],[2x]\ge[x]+\frac{[2x]}2,...,[nx]\ge[x]+\frac{[2x]}2+...+\frac{[nx]}n:$

$$[x]+[2x]+...+[nx]\ge n[x]+(n-1)\frac{[2x]}2+...+\frac{[nx]}n.$$

Прибавим к обеим частям сумму $[x]+[2x]+...+[nx]$ и воспользуемся неравенством $[a+b]\ge[a]+[b]$:

$$n[(n+1)x]\ge\left([x]+[nx]\right)+\left([2x]+[(n-1)x]\right)...+\left([nx]+[x]\right)\ge (n+1)\left([x]+\frac{[2x]}2+...+\frac{[nx]}n\right),$$

$$\frac n{n+1}[(n+1)x]\ge[x]+\frac{[2x]}2+...+\frac{[nx]}n,$$

$$[(n+1)x]\ge[x]+\frac{[2x]}2+...+\frac{[nx]}n+\frac{[(n+1)x]}{n+1}.$$

[rsoldo]

If you need... ( Hermite's ident. )

$$\lfloor nx\rfloor=\sum_{i=0}^{n-1}\left\lfloor x+\frac in\right\rfloor$$