Questions tagged [a.m.-g.m.-inequality]

For questions about proving and manipulating the AM-GM inequality. To be used necessarily with the [inequality] tag.

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Find the maximum value of $ab+bc+cd+de+ef+fa,$ given that $a+b+c+d+e+f=1$

After looking at this post, I framed the following question. If $a,b,c,d,e,f$ are positive real numbers such that $a+b+c+d+e+f=1$, then find the minimum and maximum value of $ab+bc+cd+de+ef+fa$ My ...
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4 votes
1 answer
182 views

Prove or disprove that the inequality is valid if $x,y,z,u$ are positive numbers and $x+y+z+u=2$.

Prove or disprove that the inequality $$ \dfrac{x^2}{\left(x^2+1\right)^2}+\dfrac{y^2}{\left(y^2+1\right)^2}+\dfrac{z^2}{\left(z^2+1\right)^2}+\dfrac{u^2}{\left(u^2+1\right)^2} \leq \dfrac{16}{25}$$ ...
4 votes
1 answer
85 views

If $xy+yz+zx=3$ and $x,y,z\geq0$, prove that: $\frac{1}{1+3x-p}+\frac{1}{1+3y-p}+\frac{1}{1+3z-p}\leq\frac{3}{1+2p}$

If $xy+yz+zx=3$ and $x,y,z\geq0$, prove that: $$\sum_{cyc}\frac{1}{1+3x-p}\leq\frac{3}{1+2p}$$ where $p=xyz$. *some people are not familiar with the $\sum_{cyc}$ notation, alternative would be $$\frac{...
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3 answers
159 views

JBMO-$2014$ Inequality question [duplicate]

Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that $$ {\left(a+\frac{1}{b}\right)^2}+{\left(b+\frac{1}{c}\right)^2} +{\left(c+\frac{1}{a}\right)^2}≥3(a+b+c+1)$$ My solution: By Jensen'...
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5 votes
2 answers
111 views

AM-GM & Minimization Proof [duplicate]

I want to prove that for all $x, y > 0$, $$\cfrac{x+y}{2} \geq \sqrt{xy}$$ Particularly, I want to show that the minimum of $(x+y)/2$ is exactly $\sqrt{xy}$. This is my attempt: $\textbf{Proof}$ (...
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-4 votes
1 answer
79 views

If $x_1 ,\dots , x_n$ are real numbers then $(x_1 \dots x_n )^{\frac{1}{n}} \leq \frac{x_1 + x_2 + \dots x_n}{n}$ [closed]

The algebraic mean is biger than or equal to geometric mean. It is easy to prove the case $n=2$. I tried to use induction but I guess it doesn't work. Can anybody give a proof?
1 vote
3 answers
70 views

For $x,y,z∈ℝ^{+}$,without using Hölder's inequality prove that $\frac{x}{\sqrt{x^2+8yz}}+\frac{y}{\sqrt{y^2+8xz}}+\frac{z}{\sqrt{z^2+8xy}}\geq1$.

For $x,y,z∈ℝ^{+}$, prove that $\frac{x}{\sqrt{x^2+8yz}}+\frac{y}{\sqrt{y^2+8xz}}+\frac{z}{\sqrt{z^2+8xy}}\geq1$. In this question solution used Hölder's inequality, but I am looking a solution ...
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2 votes
1 answer
124 views

Proof or references for strengthened AM-GM

This other question includes the following strengthened version of the arithmetic-mean geometric-mean inequality. \begin{equation} \label{1}\tag{1} \dfrac{a+b}{2} - \sqrt{ab} \geq \dfrac{1}{16 \max \...
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0 votes
1 answer
37 views

$\left\{(x_1,x_2)\mid \left|f(x_1,x_2)-f(0,0)\right|<\frac{1}{2}\right\}=\left\{(0,0)\right\}\cup\left\{(x_1,x_2)\mid x_1\neq\pm x_2\right\}$

I am reading "Introduction to Set Theory and Topology" (in Japanese) by Kazuo Matsuzaka. Problem 18 on p.194 Let $f$ be a function from $\mathbb{R}\times\mathbb{R}$ to $\mathbb{R}$ such ...
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2 votes
1 answer
78 views

Why does the AM-GM inequality not show $25 \csc^2(\theta) +16 \sin^2(\theta)$ has a minimum of $41$ as the graph indicates?

Let's say we have to find range of $f(\theta) = 25 \csc^2(\theta) +16 \sin^2(\theta)$ If I use $AM \ge GM$ Then $f(\theta) \ge 40$ Which tells minimum value of $f(\theta)$ will be $40$ But I checked ...
1 vote
2 answers
68 views

Prove or disprove the inequality if $a,b,c>0$, $a \geq b+c$.

Prove or disprove the inequality $$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \geq 7abc$$ if $$a,b,c>0, a \geq b+c.$$ I thought to use this evaluation: $$a^2b+b^2c+c^2a \geq 3abc.$$ So we have: $$a^2b+a^2c+b^2a+...
2 votes
2 answers
82 views

Prove or disprove that the inequality is valid if $x,y,z$ are positive numbers and $xyz=1$.

Prove or disprove that the inequality $$\dfrac{1}{\sqrt{1+x}}+\dfrac{1}{\sqrt{1+y}}+\dfrac{1}{\sqrt{1+z}} \geq 1$$ is valid if $x,y,z$ are positive numbers and $$xyz=1.$$ My solution is: Let $$x=\...
1 vote
1 answer
90 views

Show that $x^{\frac1x}<1.5$ for $x \in \mathbb R$

Well, I've proven that : $$x^{\frac1x}<1.5$$ for $x \in \mathbb R^+$, or more specifically I've shown that the maximum value of the expression is $1.44...$ for $x = e$. But I used calculus (finding ...
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2 votes
1 answer
120 views

Proof of a tighter inequality than Cauchy-Schwarz inequality

A few days ago, I found this post which discuss about some tighter versions of Cauchy-Schwarz (or some might prefer the name AM-GM) inequality. User @Michael Rozenberg proposed a very interesting ...
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0 answers
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Prove that the inequality is valid if $a,b,c,d$ are positive numbers. [duplicate]

Is given that $a,b,c,d$ are positive numbers. We have to prove that $$\dfrac{a}{b+c}+\dfrac{b}{c+d}+\dfrac{c}{a+d}+\dfrac{d}{a+b} \geq 2.$$ What have I done? I used the substitution $$a=\dfrac{1}{x},b=...
7 votes
3 answers
148 views

To prove $1^1\cdot2^2\cdot 3^3...\cdot n^n<(\frac{2n+1}{3})^{\frac{n(n+1)}{2}} $

So we have to prove the following for $n\in N $ $$1^1\cdot 2^2\cdot 3^3...\cdot n^n<\left(\frac{2n+1}{3}\right)^{\frac{n(n+1)}{2}} $$ So I used concept of weighted means (arithmetic and geometric) ...
0 votes
1 answer
31 views

Does the AM-GM inequality not hold if there is a variable on the GM side?

To find the minimum value of $x+\dfrac{1}{x}$, $(x>0)$ we can use the AM-GM inequality to say that $$\dfrac{x+\dfrac{1}{x}}{2}\geq \sqrt{x\cdot \dfrac{1}{x}}$$ or $$x+\dfrac{1}{x}\geq 2$$ The ...
0 votes
1 answer
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Prove that the inequality is valid if $x,y,z$ are positive numbers and $xyz=1.$

Is given that $x,y,z$ are positive numbers and $xyz=1$, prove that $$\dfrac{\dfrac{1}{x}}{\sqrt{z^2+1}}+\dfrac{\dfrac{1}{y}}{\sqrt{x^2+1}}+\dfrac{\dfrac{1}{z}} {\sqrt{y^2+1}}>\sqrt{2}.$$ What have ...
1 vote
2 answers
72 views

Help w/$(\frac{a}{b})^4+(\frac{b}{c})^4+(\frac{c}{d})^4+(\frac{d}{e})^4+(\frac{e}{a})^4\ge\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{e}{d}+\frac{a}{e}$ [closed]

How exactly do I solve this problem? (Source: 1984 British Math Olympiad #3 part II) \begin{equation*} \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}...
1 vote
0 answers
25 views

Sequence of geometric mean subtracted by arithmetic mean

Let $a_1,a_2,a_3,\dots$ be a sequence of positive numbers. Define $$G_n=\sqrt[n]{a_1a_2\dots a_n}~\text{and}~A_n=\frac{a_1+\dots+a_n}{n}.$$ We are supposed to use the result $$u^av^b\leq au+bv \tag{$*$...
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1 vote
0 answers
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Inequality related to AM-GM?

Let $a$ and $b$ be two positive numbers such that $a+b=1$. I am supposed to show that $u^av^b\leq au+bv$ for all positive $u$ and $v$. It is known that $\ln x \leq x$ for all positive $x$, so I ...
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1 vote
0 answers
97 views

prove that inequality holds for all reals if and only if $|k| \leq 2n$

Prove that $$x_1^2+x_2^2+\ldots+x_{2n}^2+k \cdot x_1 \cdot x_2 \cdot \ldots \cdot x_{2n}\geq 0$$ for all reals $x_1, x_2, \ldots, x_{2n}$ if and only if $|k| \leq 2n$, where $n$ and $k$ integers, $n &...
0 votes
2 answers
85 views

Let $a,b,c$ be positive real numbers such that $\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=1.$ Prove that $abc \le \frac{1}{8}$.

Let $a,b,c$ be positive real numbers such that $$\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=1.$$ Prove that $abc \le \frac{1}{8}$. So I tried this problem and came up with a solution, but I'm afraid ...
3 votes
3 answers
66 views

Cyclic inequality. $\sum_{cyc} \frac{x+2y}{\sqrt{z(x+2y+3z)}}\geq \frac{3\sqrt{6}}{2}$

Let $x,y,z>0$. Show that $$\sum_{cyc} \frac{x+2y}{\sqrt{z(x+2y+3z)}}\geq \frac{3\sqrt{6}}{2}$$ Equality case is for $x=y=z$. A hint I have is that I have to amplify with something convenient, and ...
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3 votes
0 answers
95 views

Maximum value of a variable in system of two equations

Consider the system of equations $$ 3a + 2b + c+ d =14 $$ $$ a^{2} + b^{2} + c^{2} + d^{2} =14.$$ Is there any way to find maximum value of $d$ using AM-GM inequality. I am not even able to think, how ...
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1 vote
1 answer
46 views

AM - GM inequality with Bernoulli's inequality

I am stuck on this proof: The AM-GM Inequality is Equivalent to the Bernoulli Inequality. I get the first part with the Bernoulli Inequality, but how did he get from $\frac{A_n}{A_{n-1}}\ge \frac{x_n}{...
3 votes
1 answer
97 views

For which $(n,m)$ holds $x^ny^m + x^my^n \le 2$, where $x+y = 2$?

For which $(n,m)$ holds $x^ny^m + x^my^n \le 2$, where $x+y = 2$ and $x,y \ge 0$? It is $n,m \ge 0$ where $n$ and $m$ do not have to be integers. The conjectured answer is: for $|n-m | \le \...
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-1 votes
1 answer
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Is this proof using AM-GM valid?

I was reading about the AM-GM inequality at brilliant.org. In one of the examples, it proves that: $(a^2+1)(b^2+1)\ge 4ab , ab\in\mathbb{R^+}$ But I thought applying the AM-GM inequality to a term ...
3 votes
1 answer
83 views

Prove inequality for positive real numbers $x,y,z$

For positive, real numbers $x,y,z$ prove that $$\frac{2(x+y+z)}{3}+\frac{3xyz}{xy+yz+zx}\ge3\sqrt[3]{xyz}$$ It's pretty obvious that the solution will have something to do with the $A.M-G.M$ ...
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1 vote
1 answer
67 views

Let $x,y,z\in\mathbb{R}^{+}$ such that $x+y+z = 1.$ T/F: $\frac{x^2 y^2}{(x+y)^2}+\frac{y^2 z^2}{(y+z)^2}+\frac{z^2 x^2}{(z+x)^2}\geq\frac{9}{4}xyz$

In Springer's "Inequalities" book, Exercise $2.5$ is the following: Let $x,y,z\in\mathbb{R}^{+}\ $ such that $\ x+y+z = 1.\ $ Then: $$ xy + yz + zx \geq 9xyz.\qquad (1)$$ Proof: Applying AM $...
0 votes
0 answers
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Please help me on how to prove the equality case of the inequality

goal: for n=2,3,4.... prove that given x1x2...xn=1, then x1+x2+x3+...+xn=n iff x1=x2=x3=...=xn=1 I tried to prove by induction. While the "if" direction is obvious, but I am kind of stuck in ...
1 vote
0 answers
25 views

$ab+bc+ca \le 4{\sqrt 3}\Delta$ for a triangle with sides $a$, $b$ and $c$ with area $\Delta$ [duplicate]

If $a$, $b$ and $c$ are the sides of a triangle with area $\Delta$, prove that $ab + bc + ca \le 4\sqrt3\Delta$ and prove that the equality holds iff the triangle is equilateral. I tried to approch ...
0 votes
2 answers
101 views

Possible flaw in AM GM inequality

AM GM inequality states that for real positive numbers x,y $$x + y \geq 2(xy)^{1/2}$$ You would get the least value for x = y We can also write this as, $$\frac x3 + \frac x3 + \frac x3 + y \geq 4\...
3 votes
2 answers
154 views

Find the minimum of $xy+yz+zx+\frac1x+\frac2y+\frac5z$ for $x, y, z > 0$

Let $x,y,z > 0$. What is the smallest value of the below expression? \begin{align*} xy+yz+zx+\frac1x+\frac2y+\frac5z \end{align*} I tried with Lagrange multiplier but I have not any constraint.
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-1 votes
1 answer
71 views

Show an inequality in three variables

Problem : Let $a,b,c>0$ then we have : $$\left[4-\frac{3\left(\sqrt{ac}+\sqrt{bc}+\sqrt{ab}\right)}{a+b+c}\right]^{\frac{3}{2}}abc-\left(\frac{16}{9\sqrt{3}}\right)^{\frac{1}{2}}\left(\frac{a+b+c}{...
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5 votes
2 answers
61 views

I can't come up with the intended solution to $\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab>18$

I was going trough some easy algebra problems when I encountered $$ \frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab>18. $$ As you can see the problem is easily solvable with AM > GM I fairly ...
2 votes
2 answers
109 views

Proving $30x + 3y^2 + \frac{2z^3}{9} + 36 (\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}) \ge 84$ [closed]

If $x, y, z$ are positive real numbers, prove that $$30x + 3y^2 + \frac{2z^3}{9} + 36 \left(\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}\right) \ge 84.$$ I genuinely have no clue on how to proceed. Is ...
1 vote
0 answers
85 views

min of $c$ in such that $\frac{\sum^n_{i=1}a_i}{n\sqrt[n]{\prod^n_{i=1}a_i}}\leq \left(\frac{\sqrt{a/b}+\sqrt{b/a}}2\right)^c$ for all $a_i\in [a,b]$. [closed]

Positive integer $n\geq 2$ is given. All $a_i\in[a,b]$($0<a<b$). What's the minimum value of $c$ such that $$\frac{\sum^n_{i=1}a_i}{n\sqrt[n]{\prod^n_{i=1}a_i}}\leq \left(\frac{\sqrt{\frac{a}b}+\...
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3 votes
1 answer
96 views

Prove that: $\sum_{1\le j\le n}{\frac{x_j}{\sum_{1\le i\le n} x_i-x_j}}\ge\frac{n}{n-1}$

$\color{red}{\textbf{Problem:}}$ Let, $x_i>0,1\le i\le n$, then Prove that: $$\sum_{1\le j\le n}{\frac{x_j}{\sum_{1\le i\le n} x_i-x_j}}\ge\frac{n}{n-1}$$ $\color{red}{\textbf{Proof:}}$ Using AM-...
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5 votes
3 answers
280 views

Finding the minimum value of the expression $xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2$

From R. D. Sharma's Objective Mathematics, Given that $x + y + z = 1$, ($x,y,z$ are positive real numbers) find the minimum value of $$A = xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2$$ My attempt: By A.M.-G.M....
1 vote
1 answer
50 views

Geometric-arithmetic mean inequality applied to eigenvalues

I've applied the arithmetic-geometric mean inequality to the eigenvalues of a positive definite matrix $X$, so $det(X)^{1/n}≤tr(x)/n$. Now I would like to show when equality holds. I already found out ...
1 vote
2 answers
71 views

If a, b, c are positive real numbers such that $a^2+ b^2+ c^2 = 1$ $( \frac{1}{a} +\frac{1}{b} +\frac{1}{c}) +a +b +c \geq 4\sqrt{3}$

If a, b, c are positive real numbers such that $a^2+ b^2+ c^2 = 1$ Show that: $$\frac{1}{a} +\frac{1}{b} +\frac{1}{c}+a +b +c \geq 4\sqrt{3}.$$ My attempt: First , I used Holder's : $$\frac{1}{a}+\...
2 votes
3 answers
117 views

$a+b+c=1$, prove $(1+a)(1+b)(1+c) \ge 8(1-a)(1-b)(1-c) $, without AM-GM

$a,b,c \in \mathbb{R}^{+}$, if $a+b+c=1$, prove that $$ (1+a)(1+b)(1+c) \ge 8(1-a)(1-b)(1-c) $$ but without AM-GM as the only tool. Collecting data: Since $a+b+c=1$, we cannot have one of the ...
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0 votes
0 answers
40 views

if a,b,c>0 and $abc=1$ prove that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq a+b+c$ [duplicate]

problem: a,b,c>0 and $abc=1$ prove that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq a+b+c$ my attempt: $LHS=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=(a+b+c) \frac{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}{a+b+...
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4 votes
1 answer
547 views

Weighted AM-GM Inequality

I was tried to learn some tools on inequalities, and I learn the "Weighted AM-GM Inequality" in a small book: in this book there is an exercise and a solution, I did my attempt and it was ...
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0 votes
2 answers
81 views

prove that :$(1+\frac{1}{n})^k<1+\frac{k}{n}+\frac{k^2}{2n^2}$.

if $ n,k \in \mathbb{N^*}$ and $(k-1)^2<n$ prove that :$(1+\frac{1}{n})^k<1+\frac{k}{n}+\frac{k^2}{2n^2}$. my attempt: $(1+\frac{1}{n})^k+1=(1+\frac{1}{n}).(1+\frac{1}{n})^{k-1}+1\leq \sqrt{(1+\...
user avatar
0 votes
2 answers
93 views

Let $f:\mathbb R^+\to\mathbb R$ be a continuous function satisfied $f(a)+f(b)\ge f(2\sqrt{ab})$ for all $a,b>0$ , is $f$ differentiable? [closed]

Let $f:\mathbb R^+\to\mathbb R$ be a continuous function satisfied $f(a)+f(b)\ge f(2\sqrt{ab})$ for all $a,b>0$ , is $f$ differentiable? Morever, if for all $a_1,a_2,\cdots,a_n>0$ there holds $$\...
0 votes
0 answers
78 views

Prove $ \frac{(c-a)^{2}}{6c} \le \frac{a+b+c}{3} - \frac{3}{ 1/a + 1/b+ 1/c}$

Given real numbers $c \ge b \ge a>0$, prove that $$ \frac{(c-a)^{2}}{6c} \le \frac{a+b+c}{3} - \frac{3}{ \frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$$ *using well-known inequality Other solution ...
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1 vote
1 answer
72 views

An Application of Cauchy Schwarz - AM-GM in Discrete Probability Measures

Suppose $p_m \geq 0$ and $\sum_{m \in \mathbf{Z}} p_m =1 .$ That is $p$ is a probability measure on integers. Then how can I show (is it true) that $$ \sum_{m \in \mathbf{Z}} (p_m + p_{m+1} + p_{m+2})^...
2 votes
0 answers
58 views

Finding a limit using AM-GM?

Doing some calculus papers before going university and I found this question: Find $\lim_{x\to\infty} \left[\frac{1}{3} \left(3^\frac{1}{x} + 8^\frac{1}{x} + 9^\frac{1}{x} \right)\right]^x$ My ...

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