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
drmath's user avatar
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3 votes
1 answer
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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 ...
 Alice Malinova's user avatar
1 vote
3 answers
88 views

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 $$ 2 \left(a^2+b^2+c^2\right)+\dfrac{4}{3}\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right) \geq 5 $$ ...
 Alice Malinova's user avatar
1 vote
3 answers
89 views

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 ...
 Alice Malinova's user avatar
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2 answers
80 views

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 \...
 Alice Malinova's user avatar
1 vote
0 answers
23 views

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}...
Ninga Nijx's user avatar
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2 answers
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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}+\...
 Alice Malinova's user avatar
1 vote
2 answers
103 views

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+ ...
Ninga Nijx's user avatar
1 vote
0 answers
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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 ...
aqualubix's user avatar
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2 votes
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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 ...
aarbee's user avatar
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3 votes
1 answer
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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, ...
Ninga Nijx's user avatar
0 votes
0 answers
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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 ...
user775349's user avatar
0 votes
2 answers
76 views

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\...
user31459's user avatar
  • 366
6 votes
6 answers
111 views

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}...
Michael's user avatar
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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. ...
Teufel's user avatar
  • 63
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0 answers
27 views

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+...
Ninga Nijx's user avatar
3 votes
2 answers
129 views

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 ...
IraeVid's user avatar
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1 vote
3 answers
65 views

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. ...
Will Ford's user avatar
1 vote
1 answer
129 views

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}-\...
 Alice Malinova's user avatar
1 vote
2 answers
85 views

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+...
 Alice Malinova's user avatar
4 votes
0 answers
108 views

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 ...
jasper's user avatar
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0 votes
1 answer
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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\...
Adam Rubinson's user avatar
0 votes
1 answer
68 views

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} = \...
user avatar
3 votes
1 answer
95 views

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\...
Adam Rubinson's user avatar
0 votes
1 answer
33 views

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)}^...
jessexknight's user avatar
2 votes
1 answer
80 views

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+...
hamma04's user avatar
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2 votes
1 answer
53 views

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 $$...
OpenLearner's user avatar
1 vote
0 answers
26 views

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{...
Subham Karmakar's user avatar
0 votes
3 answers
100 views

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)\...
diana's user avatar
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0 votes
1 answer
32 views

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 ...
wonderman's user avatar
1 vote
0 answers
122 views

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 ...
aarbee's user avatar
  • 7,398
4 votes
1 answer
220 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 $$ \dfrac{x^2}{\left(x^2+1\right)^2}+\dfrac{y^2}{\left(y^2+1\right)^2}+\dfrac{z^2}{\left(z^2+1\right)^2}+\dfrac{u^2}{\left(u^2+1\right)^2} \leq \dfrac{16}{25}$$ ...
 Alice Malinova's user avatar
4 votes
1 answer
88 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{...
Memat's user avatar
  • 469
-2 votes
3 answers
203 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'...
Rehman's user avatar
  • 199
5 votes
2 answers
119 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: $\textbf{Proof}$ (...
Dee's user avatar
  • 343
-4 votes
1 answer
86 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?
Abolfazl Alam's user avatar
1 vote
3 answers
94 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 ...
user avatar
3 votes
1 answer
142 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 \...
JoshuaZ's user avatar
  • 1,673
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 ...
tchappy ha's user avatar
  • 8,046
2 votes
1 answer
91 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 ...
OpenLearner's user avatar
2 votes
2 answers
78 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+...
 Alice Malinova's user avatar
3 votes
2 answers
95 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=\...
 Alice Malinova's user avatar
1 vote
1 answer
91 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 ...
D S's user avatar
  • 2,367
2 votes
1 answer
142 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 ...
Bakkune's user avatar
  • 151
7 votes
3 answers
163 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) ...
Kshitij Kumar's user avatar
0 votes
1 answer
38 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 ...
Sagewizard88's user avatar
1 vote
1 answer
83 views

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 ...
 Alice Malinova's user avatar
1 vote
2 answers
75 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}...
CrazyVideoGamer's user avatar
1 vote
0 answers
26 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{$*$...
KHOOS's user avatar
  • 331
1 vote
0 answers
53 views

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 ...
KHOOS's user avatar
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