Following on from an earlier question, and in search of a conceptual insight, I asked myself: Given real numbers $a \ge b \ge c \ge d \ge 0 \tag{1}$ Prove that $a^n – b^n + c^n – d^n \ge (a – b + c – d)^n \text{ for all } \underline{n \in \mathbb{R}} \tag{2}$ First, […]

I’m trying to figure out why $b^n – a^n < (b – a)nb^{n-1}$. Using just algebra, we can calculate $ (b – a)(b^{n-1} + b^{n-2}a + \ldots + ba^{n-2} + a^{n-1}) $ $ = (b^n + b^{n-1}a + \ldots + b^{2}a^{n-2} + ba^{n-1}) – (b^{n-1}a + b^{n-2}a^2 + \ldots + ba^{n-1} + a^{n-1}) $ $ […]

Positive real $x,y,z$ such that $$\frac1{1+x}+\frac1{1+y}+\frac1{1+z}=2.$$ Prove inequality $$8xyz<1$$ My work so far: I tried rearrangement and AM-GM but fail.

I need your support. Suppose I am performing an NTT in a finite field $GF(p)$. I assume it contains the needed primitive root of unity. I am using it to compute the convolution of two vectors of length $n=2^m, m\in \mathbb{N}$. As usual, I double the length of the vectors to $2n=2^{m+1}$ to make the […]

Let $ a_{1}, \dots a_{m} \in \mathbb{R}^{n} $ such that $ ||a_{i}-a_{j}||=1 \, \forall i \neq j $, where $ || \cdot || $ denotes the usual norm on $ \mathbb{R}^{n} $. Prove that $ m \leq n+1 $. I did it for the easy case when $ n=1 $ by explicit computation using the […]

Good morning everyone! By the arithmetic-geometric mean inequality, we all know that a suitable lower bound for the quantity $$\frac{a}{b} + \frac{b}{a}$$ is $2$. Now my question is: Will this quantity ever be bounded from above, if we allow $a$ and $b$ to be in $\mathbb{R}$? If the answer is NO, under what conditions on […]

Let $(u_n)$ be a sequence defined by: $$\begin{equation} \left\{ u_0 \geq 0 \\ \forall n \in \mathbb{N}^*, u_n = \sqrt{n+u_{n-1}} \right. \end{equation}$$ I have to prove that : $$u_n \leq n + \frac{u_0}{2^n}$$ I don’t really know where I should start to prove this… Can someone give me an hint ?

As in the title. Prove the inequality $$\frac{a^8+b^8+c^8}{a^3b^3c^3}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$ for $a,b,c>0$. Thsi inequality can be proved in a pretty straightforward manner utilizing the Muirhead’s inequality, yet I ought to prove it using the rearrangement inequality. I can’t however figure out any suitable sequences and permutations of them.

Let $a_{1}, a_{2}, \ldots, a_{n}$ and $k \geq 1$. Prove that (using Chebyshev’s inequality): $$\large \sqrt[k]{\frac{a_{1}^{k}+a_{2}^{k}+\ldots +a_{n}^{k}}{n}}\geq \frac{a_{1}+a_{2}+\ldots + a_{n}}{n}.$$ I think I have a (partial) solution but I don’t know if what I obtained really help me. So: $$\sqrt[k]{\frac{a_{1}^{k}+a_{2}^{k}+\ldots +a_{n}^{k}}{n}}\geq \frac{a_{1}+a_{2}+\ldots + a_{n}}{n} \Leftrightarrow \frac{a_{1}^{k}+\ldots+a_{n}^{k}}{n} \geq \left(\frac{a_{1}+\ldots +a_{n}}{n}\right)^{k}.$$ Now we have to prove that […]

let $a,b,c>0$, and such $$a^2+b^2+c^2=3$$ show that $$a^3b+b^3c+c^3a+a^3b^3+b^3c^3+c^3a^3\le 6\tag{2}$$ I know this famous inequality( creat by valsie) $$(a^2+b^2+c^2)^2\ge 3(a^3b+b^3c+c^3a)$$ this inequality if and only if $$a=b=c, or,a:b:c=\sin^2{\dfrac{\pi}{7}}:\sin^2{\dfrac{2\pi}{7}}:\sin^2{\dfrac{3\pi}{7}}$$ can see: Vasc inequality solution so $$(a^3b+b^3c+c^3a)\le \dfrac{1}{3}(a^2+b^2+c^2)^2=3$$ so we only prove $$a^3b^3+b^3c^3+a^3c^3\le 3 \tag{1}$$ but this (1) inequality is not true.such as $$a=\dfrac{5}{\sqrt{17}},b=\dfrac{5}{\sqrt{17}},c=\dfrac{1}{\sqrt{17}}$$ but $$a^3b^3+c^3a^3+b^3c^3\approx 3.23$$ see […]

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