Inequality with roots

Let a, b, c be positive real numbers. Prove that



We apply the AM – GM inequality, thus:

    \begin{align*} \frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}} &=\frac{a}{b}+\frac{1}{2}\sqrt{\frac{b}{c}}+\frac{1}{2}\sqrt{\frac{b}{c}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}} \\ &\geq 6\sqrt[6]{\frac{a}{b}\frac{b}{4c}\frac{c}{27a}} \\ &=\frac{6}{\sqrt[6]{4\cdot 27}} \end{align*}

Hence it suffices to prove that 3>\sqrt[6]{4\cdot 27} which holds because it is equivalent to 3^6> 4\cdot 27.

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On permutation

Let x_1, x_2, \dots, x_k be vectors of m – dimensional Euclidean space such that x_1+x_2+\cdots+x_k=0. Prove that there exists a permutation \pi of the integers \{1, 2, \dots,k\} such that

    \[\left \| \sum_{i=1}^{n} x_{\pi(i)} \right \|\leq\left ( \sum_{i=1}^{k} \left \| x_i \right \|^2 \right )^{1/2}\]

for each n=1,2, \dots, k.


We define \pi inductively. Set \pi(1)=1.Assume \pi is defined for i=1, 2, \dots, n and also

(1)   \begin{equation*} \left \| \sum_{i=1}^{n} x_{\pi(i)} \right \|^2 \leq \sum_{i=1}^{n} \left \| x_{\pi(i)} \right \|^2 \end{equation*}

Note that (1) is true for n=1. We choose \pi(n+1) in a way that (1) is fulfilled for n+1 instead of n. Set y=\sum \limits_{i=1}^{n} x_{\pi(i)} and A=\{1,2, \dots, k\} \setminus \{\pi(i) : i=1,2, \dots, n\}. Assume that (y, x_r)>0 for all r \in A.  Then \left ( y , \sum \limits_{r \in A} x_r \right )>0 and in view of y + \sum \limits_{r \in A} x_r =0 ones gets -(y, y)>0 which is impossible. Hence , there is r\in A such that

(2)   \begin{equation*} (y, x_r)\leq 0\end{equation*}

Put \pi(n+1) = r . Then using (2) and (1) we have

    \begin{align*} \left \| \sum_{i=1}^{n+1} x_{\pi(i)} \right \|^2 &= \left \| y + x_r \right \|^2 \\ &=\left \| y \right \|^2 + 2\left ( y, x_r \right ) + \left \| x_r \right \|^2 \\ &\leq \left \| y \right \|^2 + \left \| x_r \right \|^2 \\ &\leq \sum_{i=1}^{n} \left \| x_{\pi(i)} \right \|^2 + \left \| x_r \right \|^2\\ &= \sum_{i=1}^{n+1} \left \| x_{\pi(i)} \right \|^2 \end{align*}

which verifies (1) for n+1. Thus we define \pi for every n=1, 2, \dots, k. Finally from (1) we get

    \[\left \| \sum_{i=1}^{n} x_{\pi(i)} \right \|^2 \leq \sum_{i=1}^{n} \left \| x_{\pi(i)} \right \|^2 \leq \sum_{i=1}^{k} \left \| x_i \right \|^2\]

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Homomorphism and inequality

Let \mathcal{G}be a group and f:\mathcal{G} \rightarrow \mathcal{G} be a homomorphism. Prove that

    \[\left | f \left ( \mathcal{G} \right ) \right |^2 \leq \left | \mathcal{G} \right | \left | f \left ( f \left ( \mathcal{G} \right ) \right ) \right |\]


In general it holds that  |\mathcal{G}|=|\ker f| |f(\mathcal{G})| ( first isomorphism theorem ) . Taking that for granted we also have

    \[|f(\mathcal{G})| = \left|\ker f|_{f(\mathcal{G})} \right| |f(f(\mathcal{G}))|\]

and the inequality is equivelant to  \left|\ker f|_{f(\mathcal{G})} \right| \leqslant |\ker f| which is obviously true.

The exercise can also be found at .

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On linear operators

Let \alpha \in \mathbb{R} \setminus \{0\} and suppose that F ,G are linear operators from \mathbb{R}^n into \mathbb{R}^n satisfying

(1)   \begin{equation*}F\circ G - G \circ F =\alpha F \end{equation*}

  1. Show that for all k \in \mathbb{N} one has

        \[F^k \circ G - G \circ F ^k= \alpha k F^k\]

  2. Show that there exists k \geq 1 such that F^k =0.


  1. Using the assumptions we have

    \begin{aligned} F^k \circ G - G \circ F^k &= \sum_{i=1}^{k} \bigg (F^{k-i+1} \circ G \circ F^{i-1} - F^{k-i} \circ G \circ F^i \bigg ) \\ &= \sum_{i=1}^{k} \bigg( F^{k-i} \circ \left ( F \circ G- G \circ F \right ) \circ F^{i-1} \bigg)\\ &= \sum_{i=1}^{k} F^{k-i} \circ \alpha F \circ F^{i-1} \\ &= \alpha k F^k \end{aligned}

  2. Consider the linear operator \mathcal{L}(F) = F\circ G - G \circ F acting over all n \times n matrices F. It may have at most n^2 different eigenvalues. Assuming that F^k \neq 0 for every k we get that \mathcal{L} has infinitely many different eigenvalues \alpha k in view of (i). This is a contradiction.

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On the sum of inverse binomial

In this post we are discussing the sum

    \[S_n^{(m)} =\sum_{k=0}^{n} k^m \binom{n}{k}^{-1}\]

In 1947 Staver was the first to study the sum S_n^{(0)}. He observed that \displaystyle S_n^{(1)} = \frac{n}{2} S_n^{(0)}. He was , then , able to extract the recursive relation

(1)   \begin{equation*} S_{n+1}^{(0)} = \frac{n+2}{2(n+1)} S_n^{(0)} +1  \end{equation*}

He then proved a great result which is well known in literature

(2)   \begin{equation*} {\colorbox{cyan}{\color{DarkRed}\boxed{\rm{\sum_{j=0}^{n}\binom{n}{j}^{-1}=\frac{n+1}{2^{n+1}}\sum_{j=1}^{n+1}\frac{2^j}{j}}}}} \end{equation*}

Later, in 1981 Rocket combining the identity

(3)   \begin{equation*} \binom{n}{k}^{-1} = \binom{n-1}{k-1}^{-1} - \frac{n-k}{n-k+1} \binom{n}{k-1}^{-1} \end{equation*}

along with induction he was able to give another proof of (2). In 1993 Surin provided another proof using the well known integral representation of the binomial coefficient,

(4)   \begin{equation*} \binom{n}{k}^{-1} = (n+1)\int_{0}^{1} x^k \left ( 1-x \right )^{n-k} \, {\rm d}x  \end{equation*}

Since then the cases m=0 and m=1 have been studied extensively. However, Mansour generalising the idea of Sury provided a theorem which states the following:

Theorem [Mansour]: Let r,n\geq k be non negative integers and f(n, k) be given by

    \[f(n, k) = \frac{(n+r)!}{n!} \int_{u_1}^{u_2} p^k(t) q^{n-k}(t) \, {\rm d}t\]

where p, q are two functions defined on [u_1, u_2]. Let \{a_n\}_{n \in \mathbb{N}} , \{b_n\}_{n \in \mathbb{N}} be two sequences and A(x) ,B(x) be their corresponding generating functions. Then,

\displaystyle \sum_{n=0}^{\infty} \left ( \sum_{k=0}^{n} f(n, k) a_k b_{n-k} \right ) x^n = \frac{\mathrm{d}^r}{\mathrm{d} x^r} \left ( x^r \int_{u_1}^{u_2} A\left ( xp(t) \right ) B(xq(t))\, {\rm d}t \right )

Proof: The proof is a standard generating type one and is left to the reader.

Using the above theorem along with equation (4) we can generate wonderful stuff. For example:

Example 1: Pick a_n=a^n and b_n=b^n. Then

\begin{aligned} \sum_{n=0}^{\infty} \left ( \sum_{k=0}^{n} a^k b^{n-k} \binom{n}{k}^{-1} \right )x^n &= \frac{\mathrm{d} }{\mathrm{d} x} \left ( x \int_{0}^{1} \frac{{\rm d}t}{\left ( 1-axt \right )\left ( 1-bx+bxt \right )} \right ) \\ &= \frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{-\log (1-ax)-\log(1-bx)}{a+b-abx} \right ) \end{aligned}

which , after a bit of transformations gives the general result

{\colorbox{cyan}{\color{DarkRed}\boxed{\sum_{k=0}^{n}a^n b^{n-k} \binom{n}{k}^{-1} =\frac{n+1}{\left ( a+b \right )\left ( \frac{1}{a}+\frac{1}{b} \right )^{n+1}} \sum_{k=1}^{n+1} \frac{\left ( a^k+b^k \right )\left ( \frac{1}{a}+ \frac{1}{b} \right )^{k}}{k}}}}}

Fabulous, isn’t it? If we set a=b=1 we get equation (2). Of course there are other applications of the above theorem. We can establish a similar equality for the sum S_n^{(1)}. Another relation that can be established by the above theorem is the following:

(5)   \begin{equation*}{\colorbox{cyan}{\color{DarkRed} \boxed{\sum_{k=0}^{n} \binom{2n}{2k}^{-1} = \frac{n(2n+1)}{2^{2n+2}} \sum_{k=0}^{2n+1} \frac{2^k}{k+1}}}} \end{equation*}

We are not gonna go into a deep analysis but the following equalities also hold:

(6)   \begin{equation*} \binom{n}{k}^{-m} =\left [\left ( n+1 \right )^m \int_{0}^{1}t^k (1-t)^{n-k}\, {\rm d}t \right ]^m \end{equation*}


(7)   \begin{equation*} \sum_{k=0}^{n} \binom{n}{k}^{-m} =(n+1)^m \sum_{k=0}^{n}\left [ \sum_{i=0}^{k} \frac{(-1)^i}{n-k+1+i} \binom{k}{i} \right ]^m  \end{equation*}

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