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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Hint to prove an inequality
some help to prove this inequality :
$$ \forall \,\, a,b,c \geq 0 \\ \frac{a}{a^2+b^2 +2} +\frac{b}{b^2+c^2 +2} + \frac{c}{c^2+a^2 +2} \leq \frac{3}{4} $$
Thank
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Prove or disprove that the inequality is true if $xyz=1$ and $xyz=8^3$.
Prove that the inequality $$\dfrac{1}{\sqrt{x+1}}+\dfrac{1}{\sqrt{y+1}}+\dfrac{1}{\sqrt{z+1}} \geq 1$$ is true if $x, y, z>0$ and
$1)$ $xyz=1$
$2)$ $xyz=8^3.$
For the first case it turns out to ...
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Prove $2(a^2+b^2+c^2)+\frac{4}{3}\sum\limits_{\mathrm{cyc}}\frac{1}{a^2+1} \geq 5$ for $ab+bc+ac=1$
The positive real numbers a, b, and c satisfy ab+bc+ac=1. Prove that the inequality $$ 2 \left(a^2+b^2+c^2\right)+\dfrac{4}{3}\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right) \geq 5 $$ ...
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Prove $\sqrt{\frac{x}{\left(x+y\right)\left(1+y\right)}}+\sqrt{\frac{y}{\left(x+y\right)\left(1+x\right)}}+\sqrt{\frac{xy}{(1+x)(1+y)}}>1$
For $x, y, z > 0$, prove that $$\sqrt{\dfrac{x}{\left(x+y\right)\left(1+y\right)}}+\sqrt{\dfrac{y}{\left(x+y\right)\left(1+x\right)}}+\sqrt{\dfrac{xy}{\left(1+x\right)\left(1+y\right)}}>1.$$
I ...
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Prove that for positive numbers x and y the inequality is true. [duplicate]
Prove that for positive numbers x and y the inequality $$ \sqrt{\dfrac{1}{x+y}}+\sqrt{\dfrac{x}{1+y}}+\sqrt{\dfrac{y}{1+x}}>2$$ is true.
I tried to use the inequality $$\dfrac{1}{\sqrt{ab}} \geq \...
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Prove $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} + \frac{3\sqrt[3]{abc}}{a+b+c} \geq 4$ [duplicate]
$$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} + \frac{3\sqrt[3]{abc}}{a+b+c} \geq 4$$
My working out:
By AM-GM; $$\frac{a}{b} + \frac{a}{b} + \frac bc \geq 3\sqrt[3]\frac{a^2}{bc} = 3\frac{a}{\sqrt[3]{abc}...
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Prove or disprove the inequality for the length of the sides of the triangle $a, b, c$.
Prove or disprove the inequality for the length of the sides of the triangle $a, b, c$ $$\left(a^2+b^2+c^2-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2\right) \left( \dfrac{1}{a^2}+\...
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Find the maximum of $\frac {xyz}{(1+x)(x+y)(y+z)(z+16)}$, given x, y, z are positive real numbers.
Below is my working: This is equivalent to the minimum of $$\frac {(1+x)(x+y)(y+z)(z+16)}{xyz}$$
$$=(1+x)\left(1+\frac{y}{x}\right)\left(1+\frac zy \right)\left( 1 + \frac {16}{z} \right)\\$$
$$= 1+ ...
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Given the sum of $k$ numbers, what is the maximum value of their product? [duplicate]
Let $x_1,x_2,\ldots,x_k\in\mathbb{N}.$ Given that $x_1+x_2+\ldots+x_k=100,$ what is the maximum value of $P=x_1x_2\ldots x_k,$ and for what value of $k$ does this maximum occur?
This is a question ...
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If $x,y,z\in\mathbb R^+$ are in Harmonic Progression, prove that $z\cdot e^{x-y}+x\cdot e^{z-y}\ge\frac{2xz}y$
If $x,y,z\in\mathbb R^+$ are in Harmonic Progression, prove that $z\cdot e^{x-y}+x\cdot e^{z-y}\ge\frac{2xz}y$
Given, $\frac1y-\frac1x=\frac1z-\frac1y\implies z(x-y)=x(y-z)$
I tried AM-GM inequality ...
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AM-GM Inequality. Let a, b, c be positive real numbers. Prove that $ \frac {a+b+c}{3} \cdot (a^2+b^2+c^2) \ge a^2b + b^2c + c^2a$.
Prove that $$ \frac {a+b+c}{3} \cdot \frac{a^2+b^2+c^2}{3} \ge \frac{a^2b + b^2c + c^2a}{3}$$.
given that a,b,c are positive real numbers.
Solve only using AM-GM.
So far I have tried expanding LHS, ...
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bounding $(|\alpha|+1)^2+(x+|\alpha|)^2$
Let $x>0$ and $\alpha\in\mathbb{R}$ be a 'parameter'. If possible, I would like to find an upper bound for the quantity $(|\alpha|+1)^2+(x+|\alpha|)^2$. All ideas are welcome. If there's an ...
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Equality in the A.M G.M inequality
Let $a_1, a_2,\cdots a_n$ be positive real numbers. The A.M-G.M inequality states that
$$(a_1a_2\cdots a_n)^\frac{1}{n}\leq\frac{a_1+a_2+\cdots +a_n}{n}$$ with equality if and only if $a_1=a_2=a_3\...
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Show that $\frac{x^2}{x+1} + \frac{(1-x)^2}{2-x} \geq \frac 13$ for $0\leq x \leq 1$
Show that $$\frac{x^2}{x+1} + \frac{(1-x)^2}{2-x} \geq \frac 1 3$$ if $0 \leq x \leq 1$.
Source: Brilliant.
I have tried the following:
$$\frac{x^2}{x+1}=(x+1)+\frac{1}{x+1} -2$$
$$\frac{(1-x)^2}{2-x}...
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Determine the minimal value of $\frac{3a}{b+c}+\frac{4b}{c+a}+\frac{5c}{a+b}$ [duplicate]
Let $a$, $b$, $c$ be positive real numbers. Determine the minimal value of
$$\frac{3a}{b+c}+\frac{4b}{c+a}+\frac{5c}{a+b}.$$
I have solved this problem using Chebyshev's and Nesbitt's inequalities.
...
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Prove $x^2+4y^2<1$ given $x^3 + y^3 = x-y$ and x and y are positive real numbers. (Using AM-GM) [duplicate]
I have tried the following and other methods, however I haven't been able to solve this. I would very much appreciate if someone can point me in the right direction on how this can be solved.
$x-y=(x+...
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Prove that $(a+b)(b+c)(c+d)(d+a)\geq(a+1)(b+1)(c+1)(d+1)$
Question: Let $a, b, c, d$ be positive real numbers satisfying $abcd=1$. Prove that $(a+b)(b+c)(c+d)(d+a)\geq(a+1)(b+1)(c+1)(d+1)$. Characterise the instances of equality.
My (failed) approach:
I ...
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Proof of inequality $x + x^{-1} - x^r - x^{-r} \geq 0$ for $x>0$ and $0<r<1$.
I am trying to prove the inequality
$$x + x^{-1} - x^r - x^{-r} \geq 0$$
for $x>0$ and $0<r<1$. I believe this to be true as it was cited in a proof that I followed, but was not proved there. ...
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Let $a, b, c$ be positive real numbers. Prove the inequality.
Let $a, b, c$ be positive real numbers. Prove that $$\dfrac{2a^2b^2c^2}{a^3b^3+b^3c^3+c^3a^3}+\dfrac{1}{3} \geq \dfrac{3abc}{a^3+b^3+c^3}.$$
My solutions is:
$$\dfrac{3abc}{a^3+b^3+c^3}-\dfrac{1}{3}-\...
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Prove that $\sqrt{x^2+y^2}+\left(2-\sqrt{2}\right) \sqrt{xy} \geq x+y$ if $x$ and $y$ are real positive numbers!
Prove that $$\sqrt{x^2+y^2}+\left(2-\sqrt{2}\right) \sqrt{xy} \geq x+y$$ if $x$ and $y$ are real positive numbers!
Since both sides of the inequality are positive, when we square it, we get $$x^2+y^2+...
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Do the $4$ Pythagorean means divide into two algebraic dualities?
I’m looking to understand the relationship between rectangular product & mean square operations -- or, equivalently, between their root operations, the geometric & quadratic means. Here’s the ...
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Is the geometric mean of two non-negative, twice-differentiable concave real functions concave?
Is the geometric mean of two non-negative,
twice-differentiable concave real functions concave?
Let us work with the following definition of concave. A twice-differentiable real function $h:[0,1]\to\...
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Prove that if $a+b+c = 1$ then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq 3+ 2\frac{a^3+b^3+c^3}{abc}$ [duplicate]
Prove that if $a+b+c = 1$ then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq 3+ 2\frac{a^3+b^3+c^3}{abc}$
What I've done so far is to note that
$$
\begin{align}
\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \...
user733335
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Finding non-trivial real numbers which satisfy all three of: $\sum_{i=1}^{n}a_i=0,\sum_{i=1}^{n}{a_i}^3=0$ and $\sum_{i=1}^{n}\lvert a_i\rvert=1.$
A question was asked recently on this site:
Find bounds on $\ \displaystyle\sum_{i=1}^{n} {a_i} ^5\ $ if $\
\displaystyle\sum_{i=1}^{n} a_i = 0,\ \sum_{i=1}^{n} {a_i}^3 = 0\ $ and $\
\displaystyle\...
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Another inequality related to (weighted) AM-GM ...
Let $w_i \in [0,1], \sum_i w_i = 1$ (i.e. weights); $\beta_i \in (0,1)$; and $N \ge 1$.
We can show that:
$$
[A]\qquad1 - \prod_i {(1 - \beta_i)}^{N\,w_i}
\quad\ge\quad
1 - \sum_i w_i {(1 - \beta_i)}^...
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Inequality proof with AM-GM
I'm new to this community. Do you think my solution is correct?
The problem is:
$$a^2+b^2+c^2+ab+bc+ca\geq6, a+b+c=3$$
My solution:
$$a^2+b^2+c^2+2ab+2bc+2ca\geq3(ab+bc+ca)\rightarrow$$
$$3\geq ab+bc+...
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Minimum value of f(θ)= a²sec²θ +b² cosec²θ using AM- GM inequality
If we take and function
$$f(θ)= \dfrac{a^2}{\cos^2\theta} + \dfrac{b^2}{\sin^2\theta}$$
And wish to find minimum value of function using AM -GM inequality
i.e. if we have two no. p & q
Then $$...
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Is there any simple way of proving triangle inequality for $d(x,y)= \dfrac{2|x-y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}, \,\,x,y\in \mathbb{R}$ [duplicate]
$d(x,y)= \dfrac{2|x-y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}, \,\,x,y\in \mathbb{R}$
Following are the approaches I took, but can't think further -
$d(x,y)= \dfrac{2|x-y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}
= \dfrac{...
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Proving $4(x^4+y^4+z^4)\geq \sum_{cyc}xy(x+y)^2\geq 4xyz(x+y+z)$, for $x,y,z\geq0$
Trying to get used to olympiad inequalities. Tried AM-GM and did not succed. Please explain. (I'm an eight grader)
Prove that for $x,y,z\geq 0$, the following inequality holds true:
$$4(x^4+y^4+z^4)\...
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Three-Variable Inequality
Prove that if $a$, $b$, and $c$ are positive real numbers, then
$$\sqrt{a^2 + ab + b^2} + \sqrt{a^2 + ac + c^2} + \sqrt{b^2 + bc + c^2} \ge \sqrt{3} (\sqrt{ab} + \sqrt{ac} + \sqrt{bc}).$$
When does ...
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Find the maximum value of $ab+bc+cd+de+ef+fa,$ given that $a+b+c+d+e+f=1$
After looking at this post, I framed the following question.
If $a,b,c,d,e,f$ are positive real numbers such that $a+b+c+d+e+f=1$, then find the minimum and maximum value of $ab+bc+cd+de+ef+fa$
My ...
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Prove or disprove that the inequality is valid if $x,y,z,u$ are positive numbers and $x+y+z+u=2$.
Prove or disprove that the inequality $$ \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}$$ ...
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If $xy+yz+zx=3$ and $x,y,z\geq0$, prove that: $\frac{1}{1+3x-p}+\frac{1}{1+3y-p}+\frac{1}{1+3z-p}\leq\frac{3}{1+2p}$
If $xy+yz+zx=3$ and $x,y,z\geq0$, prove that:
$$\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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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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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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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?
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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 ...
user1135351
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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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$\left\{(x_1,x_2)\mid \left|f(x_1,x_2)-f(0,0)\right|<\frac{1}{2}\right\}=\left\{(0,0)\right\}\cup\left\{(x_1,x_2)\mid x_1\neq\pm x_2\right\}$
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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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 ...
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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:
$$a^2b+a^2c+b^2a+...
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Prove or disprove that the inequality is valid if $x,y,z$ are positive numbers and $xyz=1$.
Prove or disprove that the inequality
$$\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=\...
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1
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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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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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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) ...
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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 ...
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1
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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 ...
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2
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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}...
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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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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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