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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1answer
41 views

Proof of inequality : $\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{c^3+a^3+abc}\le \frac{1}{abc}$

Can we use AM-GM inequality proof method? I need a hint for proof. Similar question: To prove $\sum_{cyc}\frac{1}{a^3+b^3+abc} \le \frac{1}{abc}$ If $a,b,c \in \mathbb R$ and $a,b,c > 0$, then $$ ...
0
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2answers
47 views

Prove $(\frac{a}{b-c})^2+(\frac{b}{c-a})^2+(\frac{c}{a-b})^2 \geq 2$ [duplicate]

If $a, b, c$ are distinct real numbers, prove that $(\frac{a}{b-c})^2+(\frac{b}{c-a})^2+(\frac{c}{a-b})^2 \geq 2$ I thought of using AM-GM but that is surely not getting me anywhere ( Maybe some ...
1
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4answers
36 views

Limitations of Arithmetic-Geometric Inequality?

I have to find the range of function $f(x) = x+(1/x) +1$, where $x$ is positive. Now I did it with two ways which we can see below, in equations $(1)$ and $(2)$, by using the AM-GM inequality. $$\...
0
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2answers
59 views

Inequality involving AM-GM but its wierd

Let a, b, c be positive real numbers. Prove that $ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+ \frac{3\cdot\sqrt[3]{abc}}{a+b+c} \geq 4$ Ohk now i know using AM-GM that $ \frac{a}{b}+\frac{b}{c}+\frac{c}{...
-1
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3answers
97 views

Inequality question . [duplicate]

Let $a,b,c>0$ with$ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1$. Prove that $(a + 1)(b + 1)(c + 1) \geq 64$ Ohk so we are given that $abc=a+b+c$ with that now the inequality becomes $2abc+(a+b+c)+1 \...
2
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2answers
59 views

Inequality question.

Let $a,b,c>0$ with $a+b+c=1$. Show that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq 3 + 2\cdot\frac{(a^3 + b^3 + c^3)}{abc}$. Ohhhkk. So first off, $\begin{align} a^3 + b^3+ c^3 & =a^3 + b^...
0
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1answer
41 views

Question on Cauchy induction of AM GM Inequality

To prove the AM GM inequality, one would firstly prove the $2^k$ case such that $ a_1 a_2 a_3...a_{2^k} \leq \frac{a_1+ a_2+ a_3...+a_{2^k} }{2^k}$ given by induction. To prove the odd cases, the ...
2
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1answer
65 views

Find $\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \to \min$

I'm trying to solve \begin{align*} &\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \to \min \\ &x + y + z = 1 \\ &x, y, z > 0, \end{align*} using only inequalities. How can i solve it? I ...
1
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0answers
29 views

Find $W:= \frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y} \rightarrow \inf$

I've got two similar problems: Find \begin{align} &W:= \frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y} \rightarrow \inf \\ &x + y + z = 1 \\ &x, y, z \ge 0 \end{align} and \begin{align} &W:=...
1
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3answers
65 views

Olympiad Minimization Problem [duplicate]

I've been struggling to find a solution to this problem that I found in the archive of my country's Olympiad questions. I'm particularly interested in a solution that doesn't involve the use of ...
0
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1answer
42 views

How to prove inequality $(2n-1)x^{-1} + (2n+1)x^{2n-1} > \frac{(2n+1)^2}{2n}, \quad n > 2$ and $ 0 < x < 1$

How to prove the following inequality for $n > 2$ and $ 0 < x < 1$: $$(2n-1)x^{-1} + (2n+1)x^{2n-1} > \frac{(2n+1)^2}{2n}, \quad n \in \mathbb{N}, x \in \mathbb{R}$$ I tried using ...
2
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4answers
92 views

Prove $\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(a + b + c)$

Prove $\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(a + b + c)$ So, using AM-GM, or just pop out squares under square roots we can show: $$\sqrt{a^2 + ab + b^2} ...
1
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2answers
65 views

Show inequality $a\sqrt{b - 1} + b\sqrt{a - 1} \le ab$

Given numbers $a$ and $b$; $a, b \ge 1.$ I'm trying to prove $$a\sqrt{b-1} +b\sqrt{a - 1} \le ab.$$ Also conditions for turning it to equality. I tried to use AM-GM to the $(a - 1)(b - 1)$, which ...
2
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5answers
96 views

Prove that $\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$ with $a>0, b>0 , c> 0$ and $d>0.$

Prove that $\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$ with $a>0, b>0 , c> 0$ and $d>0.$ My attempt: $$\begin{align*}\left(\dfrac{b}{a}+\...
1
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1answer
58 views

A strange inequality

Let $a_1,...,a_n$, $b_1,...,b_n$, $x_1,...,x_n$ and $y_1,...,y_n$ be positive numbers. Assume that $\frac{\sum_{i} a_i x_i}{\sum_i b_i y_i} \leq \frac{\sum_{i} a_i^2 x_i}{\sum_i b_i^2 y_i}=1$. I ...
2
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0answers
53 views

Refinement inequality of : $\sqrt{x}+x^{\frac{x}{x+1}}\geq x+1$

Related to New bound for Am-Gm of 2 variables we have : Let $x\geq 5$ be a real number then we have : $$\sqrt{x}+x^{\frac{x}{x+1}}\geq \frac{x^2+1}{x+1}\Bigg(\frac{x^{\frac{x}{x+1}}}{x^{\...
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1answer
21 views

Maximize area of isosceles triangle with given median

Given icoscales triangle with sides $a, b, c, a = b$; median performed to the side of triangle, say, to b, denoted as $m_b.$ Note: I need to maximize area of triangle. I need to solve it using ...
0
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2answers
35 views

Solve extremal problem with two constraints using AM-GM

I'm trying to solve this extremal problem: \begin{align} &xy + yz \rightarrow \max \\ &x^2 + y^2 = 2 \\ &y + z = 2 \end{align} So, using AM-GM inequality: \begin{align} (\sqrt{xy})^2 +...
3
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3answers
57 views

Using AM-GM to show that if $\{a_i\},\{f_i\}$ are positive sequences s.t. $\sum a_i=\infty$ and $f_i\to f>0$, then $(\sum f_ia_i)/(\sum a_i)\to f$

This is from DJH Garling's book, Inequalities: A Journey into Linear Analysis Suppose $\left\{a_i\right\}$ and $\left\{f_i\right\}$ are positive sequences such that: $$\sum^\infty_{i=1}a_i=\infty$$ ...
2
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2answers
78 views

Olympiad Inequality Question

I am new to the Olympiad-style questions and I hope someone could correct my proof for this question as I do not have the answer for it. Please leave some constructive criticism if possible so I could ...
7
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2answers
91 views

Prove $\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\geqq 1$

For $x,y,z>0$ and $\sqrt{x} +\sqrt{y} +\sqrt{z} =1.$ Prove that$:$ $$\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\geq 1$$ My solution$:$ Let $x=...
2
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3answers
72 views

Prove $(a+b+c)^3 (a+b-c)(b+c-a)(c+a-b) \leqq 27a^2 b^ 2 c^2$

For $a,b,c>0$$,$ prove$:$ $$(a+b+c)^3 (a+b-c)(b+c-a)(c+a-b) \leqq 27a^2 b^ 2 c^2$$ My proof by S-S method$,$ see here. Another proof by $pqr$ method$:$ Let $p=a+b+c,\,q=ab+bc+ca,\, r=abc.$ This ...
0
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2answers
81 views

$AM-GM$ for $(a + \frac{1}{a})^2 + (b + \frac{1}{b})^2 \geqslant \frac{25}{2}$ [duplicate]

Let $a, b \in \mathbb{R_+}$ such that, $a + b = 1$. Show that $$(a + \frac{1}{a})^2 + (b + \frac{1}{b})^2 \geqslant \frac{25}{2}.$$ This was on a problem set for practising $AM-GM$ and I couldn’t ...
1
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0answers
38 views

Prove $\sum \sqrt{3a+\frac{1}{b}} \geq 6$ [duplicate]

For $a,b,c>0$ and $a+b+c=3$. Prove that$:$ $$\sqrt{3a+\frac{1}{b}} +\sqrt{3b+\frac{1}{c}} +\sqrt{3c+\frac{1}{a}} \geqq 6$$ My work$:$ This inequality equivalent to$:$ $$3(a+b+c) +\frac{1}{a}+\...
1
vote
1answer
59 views

Inequality $\sum_{cyc}\frac{a}{2a^2+a+1}\leq \frac{3}{4}$

Let $a,b,c\in\mathbb{R^+}$ such that $a+b+c=3$. Then prove that $$\sum_{cyc}\frac{a}{2a^2+a+1}\leq \frac{3}{4}$$ I tried to use tangent line method. Let $$f(x)=\frac{x}{2x^2+x+1}$$ Then $$f'(x)=\frac{...
4
votes
3answers
128 views

Prove that $\frac{1}{1 - \sqrt{ab}} + \frac{1}{1 - \sqrt{bc}} + \frac{1}{1 - \sqrt{ca}} \leq \frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$

Given that $0 < a , b , c < 1$. Prove that $\frac{1}{1 - \sqrt{ab}} + \frac{1}{1 - \sqrt{bc}} + \frac{1}{1 - \sqrt{ca}} \leq \frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$. I tried ...
0
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2answers
25 views

finding x,y,z for optimizing expression

The question is as follows: let (x,y,z) be an ordered triplet of real numbers such that x<1 ,y<2 ,z<3 and x+ y/2 + z/3 >0. for x=a ,y=b,z=c the value of expression (1-x)(2-y)(3-z)(x+y/2+z/...
3
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1answer
105 views

If $a+b+c+d=4$ Prove that $ \sqrt{\frac{a+1}{a b+1}}+\sqrt{\frac{b+1}{b c+1}}+\sqrt{\frac{c+1}{c d+1}}+\sqrt{\frac{d+1}{d a+1}} \geq 4 $

Question - Let $a, b, c, d$ be non-negative real numbers with sum 4. Prove that $ \sqrt{\frac{a+1}{a b+1}}+\sqrt{\frac{b+1}{b c+1}}+\sqrt{\frac{c+1}{c d+1}}+\sqrt{\frac{d+1}{d a+1}} \geq 4 $ My work ...
2
votes
1answer
78 views

If $x+y+z=1$ prove $ \sqrt{x+\frac{(y-z)^{2}}{12}}+\sqrt{y+\frac{(z-x)^{2}}{12}}+\sqrt{z+\frac{(x-y)^{2}}{12}} \leq \sqrt{3} $

Question - Let $x, y, z$ be non-negative real numbers with sum $1 .$ Prove that $$ \sqrt{x+\frac{(y-z)^{2}}{12}}+\sqrt{y+\frac{(z-x)^{2}}{12}}+\sqrt{z+\frac{(x-y)^{2}}{12}} \leq \sqrt{3} $$ My work -...
3
votes
2answers
75 views

Prove that $\sqrt{\frac{2 a^{2}+b c}{a^{2}+2 b c}}+\sqrt{\frac{2 b^{2}+c a}{b^{2}+2 c a}}+\sqrt{\frac{2 c^{2}+a b}{c^{2}+2 a b}} \geq 2 \sqrt{2}$

Question - Prove that for all non-negative real numbers a,b, c, we have $$ \sqrt{\frac{2 a^{2}+b c}{a^{2}+2 b c}}+\sqrt{\frac{2 b^{2}+c a}{b^{2}+2 c a}}+\sqrt{\frac{2 c^{2}+a b}{c^{2}+2 a b}} \geq 2 \...
18
votes
2answers
911 views

Is the AM-GM inequality the only obstruction for getting a specific sum and product?

This might be silly, but here it goes. Let $P,S>0$ be positive real numbers that satisfy $\frac{S}{n} \ge \sqrt[n]{P}$. Does there exist a sequence of positive real numbers $a_1,\dots,a_n$ such ...
0
votes
1answer
55 views

Proof that $\sum a_k a_{k+1} \leq \dfrac{a^2}{4} $ where $a = \sum a_k $

Let $a_k > 0$ ( $k=1,...,n$ ) nad set $a = a_1 + ... +a_n $. Prove that $$ \sum_{k=1}^{n-1} a_k a_{k+1} \leq \dfrac{a^2}{4} $$ Proof. Notice that the LHS is equivalent to $\sum_{k=1}^{n-...
3
votes
1answer
61 views

Prove that $\frac{x^{2}}{(x-y)^{2}}+\frac{y^{2}}{(y-z)^{2}}+\frac{z^{2}}{(z-x)^{2}} \geq 1$ [duplicate]

Question - Let $x, y, z$ be distinct real numbers. Prove that $$ \frac{x^{2}}{(x-y)^{2}}+\frac{y^{2}}{(y-z)^{2}}+\frac{z^{2}}{(z-x)^{2}} \geq 1 $$ My work - first i apply directly C-S and after ...
4
votes
2answers
92 views

If $a+b+c=3$ Prove that $a^{2}+b^{2}+c^{2}\geq\frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a}$

Question - Let $a, b, c$ be positive real numbers such that $a+b+c=3 .$ Prove that $$ a^{2}+b^{2}+c^{2} \geq \frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a} $$ My try - i tried putting $a+2 = x,...
0
votes
4answers
61 views

MOP 2011 inequality

If $a,b,c$ are positive integers prove that $\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +\sqrt{(a^2+c^2-ac)} +9(abc)^{1/3} \le 4(a+b+c)$ My attempt: I tried to split inequality and prove it bit by ...
0
votes
4answers
74 views

Prove that $a^{2} \sqrt{b^{2}-b c+c^{2}}+b^{2} \sqrt{c^{2}-c a+a^{2}}+c^{2} \sqrt{a^{2}-a b+b^{2}} \leq a^{3}+b^{3}+c^{3}$

Question - Let $a, b, c$ be non-negative real numbers. Prove that $$ a^{2} \sqrt{b^{2}-b c+c^{2}}+b^{2} \sqrt{c^{2}-c a+a^{2}}+c^{2} \sqrt{a^{2}-a b+b^{2}} \leq a^{3}+b^{3}+c^{3} $$ my doubt - ...
0
votes
1answer
25 views

Is this “constrained” infimum zero?

Let $s>0$ be a fixed real number. Does $$ \inf_{x,y>0,\, xy \ge \frac{1}{4},\,x-y=s}(\sqrt{x}-\sqrt{y})^2\big( (\sqrt{x}+\sqrt{y})^2-2\big)=0\,\,\,?$$ Note that by the AM-GM inequality, we ...
1
vote
1answer
74 views

Does this inequality hold with some constant factor $c>0$?

Does there exist a real number $c>0$ such that $$ (x-1)^2+(y-1)^2-2(\sqrt{xy}-1)^2\ge c\big( (x-\sqrt{xy})^2+(y-\sqrt{xy})^2 \big) \tag{*}$$ holds for every positive real numbers $x,y$ such that $...
3
votes
3answers
109 views

Prove that if $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ then $ \frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1 $

Question - Let $a, b, c$ be positive real numbers such that $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ . Prove that $$ \frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1 $$ My try $$...
0
votes
2answers
55 views

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

Question Let.a, $b, c$ be positive real numbers with sum 3 . Prove that $$ \frac{a^{2}}{a+2 b^{3}}+\frac{b^{2}}{b+2 c^{3}}+\frac{c^{2}}{c+2 a^{3}} \geq 1 $$ my doubt - by using cauchy reverse ...
1
vote
1answer
56 views

How to analyze the equation $(x-y)^2=2\big( (x+y)-2\sqrt{xy} \big)$?

Suppose that $x,y$ are positive real numbers and that $$ (x-y)^2=2\big( (x+y)-2\sqrt{xy} \big). \tag{*}$$ Then Mathematica claims that one of the following $3$ options holds: $$1. \, \, \, x=y.$$ $...
1
vote
1answer
43 views

$\frac{a}{1+b^{2} c}+\frac{b}{1+c^{2} d}+\frac{c}{1+d^{2} a}+\frac{d}{1+a^{2} b} \geq 2 $

Question - Suppose that $a, b, c, d$ are four positive real numbers with sum 4. Prove that $$ \frac{a}{1+b^{2} c}+\frac{b}{1+c^{2} d}+\frac{c}{1+d^{2} a}+\frac{d}{1+a^{2} b} \geq 2 $$ my doubt - ...
0
votes
1answer
32 views

Doubt in solution of problem in secrets in inequalities by pham kim hung

Question - Suppose that $a, b, c$ are three side-lengths of a triangle with perimeter 3. Prove that $$ \frac{1}{\sqrt{a+b-c}}+\frac{1}{\sqrt{b+c-a}}+\frac{1}{\sqrt{c+a-b}} \geq \frac{9}{a b+b c+c a} $...
0
votes
0answers
19 views

Proving $|\prod_{i=0}^n(x-x_i)|\leq\frac{n!}{4}(\frac{b-a}{n})^{n+1}$ [duplicate]

My task is to prove: $$ |\prod_{i=0}^n(x-x_i)|\leq\frac{n!}{4}(\frac{b-a}{n})^{n+1} $$ where $x_i=a+i\frac{b-a}{n}$ for $i=0,...,n$ and $x\in [a;b]$ I managed to prove that: $$ |(x-x_0)(x-x_1)|\leq \...
1
vote
2answers
48 views

Inequality $a^ab^bc^c \geq (a+b-c)^a(b+c-a)^b(c+a-b)^c$

I found a problem where now I essentially have to show the following inequality $$a^ab^bc^c \geq (a+b-c)^a(b+c-a)^b(c+a-b)^c$$ where $a,b,c$ are the sides of a triangle. I have tried a lot of ...
2
votes
1answer
41 views

Doubt in solution of APMO 1998 Inequality problem

Question - Let $a, b, c$ be positive real numbers. Prove that $$ \begin{array}{c} \left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right) \geq 2+\frac{2(x+y+z)}{\sqrt[3]{x y z}}...
0
votes
2answers
45 views

Minimize surface of cone with given volume without using derivatives

Given volume of cone equal to V, i need to minimize side surface area of given cone without using derivatives. Exactly i'm trying to use AM-GM inequality. First of all i tried to do it like it was ...
0
votes
1answer
66 views

linear programming with logarithm

max log $x_1$ + a log $x_2$ subject to $2$$x_1$+ $x_2$ $\le 8$ $x_1$+$x_2$ $\le 5$ $x_1 \ge 0$ , $x_2 \ge 0$ i need to solve and show optimal solution when a =1 but how can i find it? is it ...
0
votes
1answer
56 views

Number of ordered triplets (x,y,z) satisfying the given inequality

If $x,y,z>0$ and $x(1-y)>{1\over 4}$ , $y(1-z)>{1\over 4}$ , $z(1-x)>{1\over 4}$ then find the number of order triplets (x,y,z) satisfying the above inequaltiy. I am stuck as I know how ...
5
votes
1answer
76 views

Does $|f(\sqrt{xy})| \le |\frac{f(x)+f(y)}{2}|$ imply $f$ is a logarithm?

Let $f:\mathbb R^+ \to \mathbb R$ be a continuous function, and suppose that $$ |f(\sqrt{xy})| \le |\frac{f(x)+f(y)}{2}|, \tag{1}$$ holds for every $x,y \in \mathbb R^+$. Suppose also that $f(1)=...

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