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shorter proof of generalized mediant inequality?

A geometric proof without words : But with words, it's better. Let $V_k=\pmatrix{a_k\\b_k}$ and $\theta_k \in I$ its polar angle where $I=(0,\tfrac{\pi}{2})$ The idea is that it suffices to "...
Jean Marie's user avatar
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shorter proof of generalized mediant inequality?

Assume $$\frac{a_1}{b_1}\le\frac{a_2}{b_2}\le\dots\le\frac{a_n}{b_n}$$ and define $$N_k:=a_1+a_2+\dots+a_k\quad\text{and}\quad D_k:=b_1+b_2+\dots+b_k.$$ Using the ordinary mediant inequality, ...
Anne Bauval's user avatar
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4 votes
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Proof of Doob's Maximal Inequality for Positive Supermartingales

To show that $\mathbb{E}[X_T] \le \mathbb{E}[X_0]$, note that we have $\mathbb{E}[X_{T \wedge n}] \le \mathbb{E}[X_0]$ for all $n$ because $T \wedge n$ is a bounded stopping time. Then, since $X_{T \...
user6247850's user avatar
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Elementary central binomial coefficient estimates

You have received a lot of very nice answers but seemingly none of them has addressed your initial question. Since $$ \frac1{4^n}\binom{2n}n=\prod_{k=1}^n\frac{2k-1}{2k}:=a_n. $$ we need to prove: $$ \...
user's user avatar
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Prove that $\int_{0}^{1}f(x)^2dx\geq 4$

By Cauchy's inequality $$\displaystyle \left(\sum_{k=n}^m a_k b_k \right)^2 \le \left( \sum_{k=n}^m a_k^2\right) \left(\sum_{k=n}^m b_k^2\right)$$ or $$\left(\int_{n}^m f(x)g(x)dx\right)^2 \le \left(\...
pie's user avatar
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3 votes

challenging inequality with complex numbers

For convenience, we will use engineering notation $z^*$ for the complex conjugate of $z$ and statistical notation $\bar z$ for the arithmetic mean of the $z_i\;$ ($i=1,...,n$). First, let us assume $$\...
John Bentin's user avatar
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1 vote
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Determine all $(p,q)$ where Lorentz quasi-norm $\|\cdot\|_{L^{p,q}}$ is norm

Lorentz himself proved that $\|\cdot\|_{L^{p,q}}$ is a norm if and only if $1\leq q\leq p<\infty$ (G.G. Lorentz, On the theory of spaces $\Lambda$, Pacific J. Math. 1 (1951), 411–429). I don't know ...
Willow Wisp's user avatar
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3 votes

Inequality with logarithms and radicals of order 4

This is the general solution where we don't necessarily have to order x and y.We are allowed to logarithmize to the base x without changing the meaning of the inequality because x,y>0. logarithmize ...
Last X's user avatar
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1 vote

Looking for an existing proof for a property of triangles

This is well discussed in the Jung's theorem: https://en.wikipedia.org/wiki/Jung's_theorem
Xuefeng LIU's user avatar
1 vote
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Inequality with logarithms and radicals of order 4

In case of $x \gt y$, we have : $y \lt x \implies \log_x y \lt 1 \implies \frac{1}{\log_y x} \gt 1 \implies y^{\frac{1}{\log_y x}} \gt y$. Note here the inequality gets reversed in the second ...
Vinay Karthik's user avatar
2 votes
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Demonstrate that: $\frac{1-a^2+c^2}{5c(3a+2b\sqrt{2})}+\frac{1-b^2+a^2}{5a(3b+2c\sqrt{2})}+\frac{1-c^2+b^2}{5b(3c+2a\sqrt{2})} \geq \frac{6}{5}$

By AM-GM and Cauchy-Schwarz: $$5c(3a+2b\sqrt{2})\le \left(\frac{5c+(3a+2b\sqrt{2})}{2}\right)^2\le (3a^2+4b^2+5c^2)\left(\frac{3+2+5}{4}\right)$$ Following your idea: $$\sum \frac{3a^2+4b^2+5c^2}{5c(...
Taha Direk's user avatar
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1 vote

Prove: $\sum_{i=1}^{n-k+1} \frac{x_{n-k+1}}{X_{i}} > \sum_{i=1}^{k} \frac{x_{k}}{X_{i}}$

A counter-example with $n=5$ and $k=2$: $x_1 = 1$, $x_2 = 0.9$, $x_3=0.09$, $x_4 = 0.009$, $x_5 = 0.001$, then: $$\sum_{i=1}^4\frac{x_4}{X_i}=1.0035<\sum_{i=1}^2\frac{x_2}{X_i}=1.35. $$ The overall ...
Fred Li's user avatar
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How to prove this (geometrically correct) inequality?

By the mediant inequality, $\dfrac ab = \dfrac{(a-x_1)+x_1}{(b-x_2)+x_2}$ is between $\dfrac{a-x_1}{b-x_2}$ and $\dfrac{x_1}{x_2}$. Given $\dfrac ab < \dfrac{x_1}{x_2}$, so $$ \frac{a-x_1}{b-x_2} &...
peterwhy's user avatar
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1 vote

How to prove this (geometrically correct) inequality?

It is very easy to prove independently that (as suggested by your picture) $$\frac{a-x_1}{b-x_2}<\frac ab$$ and $$\frac{a-x_1^*}{b-x_2^*}>\frac ab.$$ For instance for the first one: $$(a-x_1)b&...
Anne Bauval's user avatar
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Need help showing $(a^p + b^p) \le (a^2 + b^2)^{p/2}$, where $p \ge 2$, and $a,b \ge 0$.

First, let's show that $$\forall q\ge 1,\forall 0\le x\le1,x^q+(1-x)^q\le 1$$ This is obvious since $x^q\le x$ and $(1-x)^q\le 1-x$, so we have $x^q+(1-x)^q\le x+1-x=1$. Next, substitute $x=\frac{a^2}{...
Fellow InstituteOfMathophile's user avatar
5 votes
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challenging inequality with complex numbers

Here's a proof using Triangle Inequality and Cauchy-Schwarz (i.) when $\sum_{k=1}^n z_k = 0$ $\sum_{k=1}^n \vert z_k\vert^2= \Big \vert \sum_{k=1}^n \overline z_k \cdot z_k - \overline z_k\cdot z\...
user8675309's user avatar
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The triangle $ABC$ is right-angled in $A$. Prove that the inequality $(1-\sin B)(1-\sin C)\leq \frac{{(\sqrt{2}-1)}^2}{2}$ holds.

In a triangle $\triangle ABC$ $$\cos A+\cos B+\cos C=1+4\sin \tfrac A2\sin\tfrac B2\sin\tfrac C2.$$ Source. Sign mistake. When $A=90^\circ,$ $$\sin\tfrac B2\sin\tfrac C2=\frac{\cos B+\cos C-1}{2\sqrt2}...
Bob Dobbs's user avatar
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4 votes
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Show $\root{-e}\of{e}<\ln2$ without a calculator

Beforehand, we will need the following "basic" facts : $(A)\;$ $2 < e < 3$; $(B)\;$ $e^x$ is a strictly monotonically increasing function; $(C)\;$ $x^{-1}$ is a strictly monotonically ...
Abezhiko's user avatar
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1 vote
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$\frac{1}{2a^2+3}+\frac{1}{2b^2+3}+\frac{1}{2c^2+3}\geq\frac{3}{5}.$

OP said that they tried to consider $ \frac{ (4a-1)(a-1)^2}{2a^2 + 3 } \geq 0$, but got stuck. To continue from there, observe that via partial fractions: $$ 0 \leq \frac{ (4a-1)(a-1)^2}{2a^2 + 3 } = \...
Calvin Lin's user avatar
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1 vote

$\frac{1}{2a^2+3}+\frac{1}{2b^2+3}+\frac{1}{2c^2+3}\geq\frac{3}{5}.$

Set $f(x) = \frac1{2x^2+3}$. Note that $f''(x) = \frac{12(2x^2-1)}{(2x^2+3)^3}$. Further $f''(x) >0$ for $x \gt \frac1{\sqrt 2}$. So the tangent to $f(x)$ at $x=1$ ie $g(x) = \frac{9-4x}{25}$ ...
Sahaj's user avatar
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2 votes
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The triangle $ABC$ is right-angled in $A$. Prove that the inequality $(1-\sin B)(1-\sin C)\leq \frac{{(\sqrt{2}-1)}^2}{2}$ holds.

WLOG the hypotenuse is 1. Let the legs be $a,b$. Denote $ab = x$. $$(1-a)(1-b) = \frac{a^2b^2}{1+a+b+ab} \le \frac{a^2b^2}{1+2\sqrt {ab}+ab} = \left(\frac{x}{1+\sqrt x}\right)^2$$ Now, consider: $$\...
D S's user avatar
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2 votes

The triangle $ABC$ is right-angled in $A$. Prove that the inequality $(1-\sin B)(1-\sin C)\leq \frac{{(\sqrt{2}-1)}^2}{2}$ holds.

As it was pointed out in other answers and comments, this amounts to determining the maximum value of $f(x)=(1-\sin x)(1-\cos x)$ for $x \in [0, \frac{\pi}{2}]$. Since $f$ is differentiable, the ...
PierreCarre's user avatar
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1 vote

challenging inequality with complex numbers

You could use the derivative mechanism with $z$ and $z^*$ (the complex conjugate) treated as independent variables. But if you aren't familiar with it, first study it to convince yourself that this is ...
Jos Bergervoet's user avatar
2 votes
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Prove that $\mathbb{P}\left(\Vert A \Vert \ge cK(\sqrt{N} + \sqrt{K} + t)\right) \le 2\exp(-t^2)$

Before starting the proof for this question, I would like to provide some following results which were proved here that we will use then in the main proof. Corollary (Covering numbers of the Euclidean ...
Tung Nguyen's user avatar
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1 vote
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An inequality about the $BMO$ space from Garfako's Modern Fourier Analysis

$\newcommand{\d}{\,\mathrm{d}}\newcommand{\avg}{\operatorname{Avg}_Q}$A direct triangle inequality bash wasn't clear to me, but working backwards and trying to squeeze as much under the integral sign ...
FShrike's user avatar
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1 vote

The triangle $ABC$ is right-angled in $A$. Prove that the inequality $(1-\sin B)(1-\sin C)\leq \frac{{(\sqrt{2}-1)}^2}{2}$ holds.

$\sin B = \dfrac{b}{a} , \sin C = \dfrac{c}{a} $, so $ (1 - \sin B) (1 - \sin C) = \dfrac{(a - b)(a - c)}{a^2}$ We can take $a = 1$ without loss of generality because we talking about angles here not ...
of course's user avatar
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$e^{xL}(x+x^3)<\frac{3\pi^2x}{L^2}$ for $x>0$ small enough and $L>0$ small enough

It isn't true for every $L>0$ because $$e^{xL}(x+x^3)=x+Lx^2+O(x^3)$$ as $x\to 0$.
uniquesolution's user avatar
0 votes

Show $\root{-e}\of{e}<\ln2$ without a calculator

Euler showed how to approximate log 2 as continued fraction, see Enestrom #606, §23. There we find $\displaystyle log(2) \approx \frac{262}{378} \approx 0.693121$ (without a calculator as requested, ...
m-stgt's user avatar
  • 217
2 votes

The triangle $ABC$ is right-angled in $A$. Prove that the inequality $(1-\sin B)(1-\sin C)\leq \frac{{(\sqrt{2}-1)}^2}{2}$ holds.

Hint : It is a right angle triangle so, $B + C = 90^{\circ}$. So, this can be re-written as $$(1-\sin B)(1-\cos B)$$ I hope you will be proceed from to get maxima at $B=45^{\circ}$ and then the final ...
Mahendra Varma's user avatar
1 vote
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Information-theoretic Inequality

Another similar approach, exploiting the convexity of the logarithmic function, we can write \begin{equation} \begin{aligned} \log\left(\mathbb{E}\left[\frac{p(y|x)}{p(y)}\right]\right) &\...
Math Lover's user avatar
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Hardy's Inequality: Problems $3.14$ and $3.15$ in Rudin's RCA

Here another approach to question 1, finding simple counter-examples. Given $a>p$, let's define this function : $$f_a(x)=x^{-1/a}\times\chi_{(0,1]}(x)$$ where $\chi_{(0,1]}(x)=1$ if $x\in(0,1]$ and ...
hdci's user avatar
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1 vote

For $a,b,c>0$ and $a+b+c=1$ prove $\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ca+2b^{2}+2b}\ge \frac{1}{ab+bc+ca}$

Due to CS: $$\sum \frac1{\left( ab+2c^2+2c \right)} \cdot \sum \left(a^2b^2 (ab+2c^2+2c) \right) \ge \left(\sum ab\right)^2.$$ Divide both parts by $\left(\sum ab\right)^3:$ $$\sum \frac1{\left( ab+2c^...
Aig's user avatar
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1 vote

For $a,b,c>0$ and $a+b+c=1$ prove $\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ca+2b^{2}+2b}\ge \frac{1}{ab+bc+ca}$

This can be solved with an application of Jensen's inequality. You want to come up with a function $f(x)$ such that $f(a)=\frac{1}{bc+2a^2+2a}$, $f(b)=\frac{1}{ac+2b^2+2b}$, and $f(c)= \frac{1}{ab+2c^...
snerd8's user avatar
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Young's inequality for scalar multiplied with absolute value of function

I believe the following holds for $x,y \in \mathbb{R}$ because $|x|^2 = x^2$ and for $y$ as awell. $$ x\cdot y \leq |x|\cdot |y| \leq \frac{1}{2\cdot \epsilon}\cdot x^2 + \frac{\epsilon}{2}\cdot y^2 $$...
d0SO'N's user avatar
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1 vote

Is there a spelling mistake or am I missing something

The double inequality $$\frac{\lfloor Nx\rfloor}N \leqslant x \leqslant \frac{\lfloor Nx\rfloor}N + \frac 1N$$ describes an interval containing the $x$ value: $$x\in\left [ \frac{\lfloor Nx\rfloor}N , ...
CiaPan's user avatar
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1 vote
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Is there a spelling mistake or am I missing something

Let $a=\frac{[Nx]}{N}$ and $b= \frac{[Nx]}{N} +\frac{1}{N}$ $\frac{[Nx]}{N} +\frac{1}{2N}=\frac{a+b}{2}$ is the midpoint of $a$ and $b$. By the first sets inequalities, we know that $x$ is between ...
Malady's user avatar
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2 votes

Prove that $a^2+b^2+c^2 \geq ab(a+b+\sqrt{ab})+cb(c+b+\sqrt{cb})+ ac(a+c+\sqrt{ac} )$

Using cyclic (not symmetric) summations: $$\begin{align} \sum a^2 &= (\sum a)(\sum a^2) \\ &= \sum a^3 + \sum a^2b+\sum ab^2 \\ &\ge \sum (ab)^{3/2} + \sum a^2b+\sum ab^2 \\ & = \sum ...
D S's user avatar
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3 votes
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Prove that $a^2+b^2+c^2 \geq ab(a+b+\sqrt{ab})+cb(c+b+\sqrt{cb})+ ac(a+c+\sqrt{ac} )$

Multiply LHS by $a+b+c=1$. You get after expansion: $$\sum a^3 + \sum(a^2b+ab^2)\ge \sum (a^2b+ab^2) +\sum a^{3/2}b^{3/2},$$ which is true due to Muirhead or just $x^2+y^2+z^2\ge xy+yz+zx.$
Aig's user avatar
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Equality in Hardy's inequality via Hölder's

I would like to suggest this approach. Let's $C_C$ be the set of continuous compact supported functions and $C_C^+$ the subset of non negative functions of $C_C$. For all $f\in C_C^+$, it's easy (...
hdci's user avatar
  • 316
4 votes
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Prove that, if $a,b,c \geq 1$, then $\frac{3ab+2c+1}{a+b}+\frac{3bc+2a+1}{b+c}+\frac{3ca+2b+1}{c+a} \geq 9$.

Use the fact that $$3ab+1\ge 2ab+2\ge 2a+2b \hspace{1cm} \text{for }a,b\ge 1$$ Then use C-S inequality $$\sum_{\text{sym}}\frac{3ab+2c+1}{a+b} \ge\sum_{\text{sym}}\frac{2a + 2b +2c}{a+b} = 2(a+b+c) \...
NN2's user avatar
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1 vote

Minimizing $2a^2 + b^2 + c^2$ given $4a + 3b + c = 7$

By application of Cauchy-Schwarz Inequality $$ \left( \sum_{i=1}^{n} a_i ^2 \right)\left( \sum_{i=1}^{n} b_i ^2 \right) \geq \left( \sum_{i=1} a_i b_i \right)^2$$ We have $$ (2a^2 + b^2 + c^2)((4/\...
Shivansh Jaiswal's user avatar
1 vote

Minimizing $2a^2 + b^2 + c^2$ given $4a + 3b + c = 7$

Hint, Let $f(a,b,c)=2a^2+b^2+c^2$ and $g(a,b,c)=4a+3b+c-7$ , Now consider $\nabla f=\lambda \nabla g$ with $g(a,b,c)=0$.
DARK's user avatar
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1 vote
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How to derive an inequality for the solution of a differential equation based on a differential inequality?

$$\begin{align} (1)&\iff \dot{y}(t) + c\cdot y(t)\leq \lambda \\ &\iff e^{ct}\left(\dot{y}(t) + c\cdot y(t) \right) \le \lambda e^{ct} \\ &\iff (e^{ct}y(t))' \le \lambda e^{ct} \\ &\...
NN2's user avatar
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Prove $1< \frac a{\sqrt{a^2+b^2}} + \frac b{\sqrt{b^2+c^2}} + \frac c{\sqrt{c^2+a^2}} \le \frac{3\sqrt2}2$

For the Right Hand Side Inequality Let $x=a^2,y=b^2,z=c^2$. Then, we just need to prove: $$\sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}} \leq \frac{3}{\sqrt{2}}$$ By C-S, we have: $$\...
Danh Trung's user avatar
1 vote

Prove the inequality knowing $x,y,z \ge 0$ and $xyz=1$ $\frac{x^5}{x^2+1}+\frac{y^5}{y^2+1}+\frac{z^5}{z^2+1}\ge\frac{3}{2}$

From the hypothesis, we can rewrite the first inequality as: $$\frac{x^4}{x+yz}+\frac{y^4}{y+zx}+\frac{z^4}{z+xy} \ge \frac{3}{2}$$ By applying Titu's lemma, we have: $$\frac{x^4}{x+yz}+\frac{y^4}{y+...
Danh Trung's user avatar
3 votes

Show by hand : $e^{e^2}>1000\phi$

Here's a very tedious calculation using repeated squaring. I did use a computer to automate and typeset, but in principle it could be performed by hand. First: $$e^z = \left(e^{\frac{z}{2^{m}}}\right)...
Claude's user avatar
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2 votes
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Equivalence of $R^m$ and $l^2$(square-summable sequence) in the proof of Grothendieck's inequality

The goal is to prove there is a universal constant $K$ such that for any $m$ and $n$, and any $m+n$ unit vectors $u_1,\dots,u_m,v_1,\dots,v_n$ in a (potentially abstract) Hilbert space $(H,\langle\...
Alex Ortiz's user avatar
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1 vote

Prove the inequality knowing $x,y,z \ge 0$ and $xyz=1$ $\frac{x^5}{x^2+1}+\frac{y^5}{y^2+1}+\frac{z^5}{z^2+1}\ge\frac{3}{2}$

No tricks requires, just standard techniques. Lemma: Apply Jensen's to show that If $ \frac{ x+y+z}{3} = A $, then $\sum \frac{ x^5 } { x^2 + 1 } \geq 3 \frac{ A^5}{A^2 + 1 }$. Proof: Work through ...
Calvin Lin's user avatar
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2 votes

Do we have $\mathbb{P}(\Vert A \Vert >\lambda) \le \mathbb{P}(\langle Ax,y\rangle > \lambda)$?

The inequality should go the other way. (I am assuming that $x,y \in \mathbb R^n$ with $\|x\|_2 = \|y_2\| = 1$, as in the definition of $\|A\|$.) If $\langle Ax,y\rangle > \lambda$ for some fixed ...
Misha Lavrov's user avatar
2 votes

Prove the inequality knowing $x,y,z \ge 0$ and $xyz=1$ $\frac{x^5}{x^2+1}+\frac{y^5}{y^2+1}+\frac{z^5}{z^2+1}\ge\frac{3}{2}$

Another solution: $$\sum \frac{x^5}{x^2+1} =\frac{1}{(xyz)^{2/3}}\cdot \sum \frac{x^5}{x^2+xyz} =$$ $$= \frac{1}{(xyz)^{2/3}}\cdot \sum \frac{x^4}{x+yz} \overset{\text{AM-GM}}\ge \frac{3}{x^2+y^2+z^2}\...
Aig's user avatar
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