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

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

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0answers
18 views

Proving $F(k)=(16X^2- 24X+18)k^2-11kX+1\geqq0$

Given that $1\leqq X\leqq k$, prove that$$F(k)=(16X^2-24X+18)k^2-11X+1\geqq0.$$ Original problem: Given that $a$, $b$, $c$ are three non−negative numbers and $a+b+c=3$, prove that$$(2+a^2)(2+b^2)(2+c^...
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1answer
37 views

Prove integral inequality :

I'm try to prove that : $\int_0^1\int_0^1\int_0^1(\frac{abc}{5}+\exp(\frac{-abc}{4}))dadbdc≤1$ Can use Cauchy Schwartz here or AM-GM Help me or hint me Thanks!
0
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2answers
47 views

How to show that $ab+bc+ca\le \frac34$

Let $a,b$ and $c$ be positive real numbers such that $(a+b)(b+c)(c+a) = 1$ , hen show that $$ab+bc+ca\le \frac34$$ I believe I need to use AM-GM inequality and use the fact $(a+b)(b+c)(c+a) = 1$ ...
0
votes
3answers
79 views

Prove inequality $1+ a^2 + b^2+ c^2+ 4\,abc \geq a + b+ c+ ab+ bc+ ca$

Given that $a,\,b,\,c$ are $3$ non$-$negatve numbers$,$ prove$:$ $$1+ a^{\,2}+ b^{\,2}+ c^{\,2}+ 4\,abc\geq a+ b+ c+ ab+ bc+ ca$$ Let$:$ $X= a+ b+ c$$,$ we have to prove$:$ $$\left ( \frac{1}{X^{\,3}}-...
7
votes
5answers
146 views

Prove that : $\frac{ab^2(b+c)}{c^2}+\frac{bc^2(a+c)}{a^2}+\frac{ca^2(b+a)}{b^2}≥2(ab+ac+bc)$ where $a,b,c>0$

Show that $$\frac{b^2(b+c)a}{c^2}+\frac{c^2(a+c)b}{a^2}+\frac{a^2(b+a)c}{b^2}≥2(ab+ac+bc)$$ where $a,b,c>0$ Can AM-GM work here? I need someone help me or hinting me please. Thanks!
1
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1answer
45 views

Proving the AM-GM Inequality with a given fact

Given $x + y + z \geq 3$ for all $(x, y, z) \in \mathbb{R}^{3}$ such that $x,y, z > 0$ provided $xyz = 1$, show that $$\frac{a_1+a_2+a_3}{3} \geq \sqrt[3]{a_1a_2a_3}$$ holds. I'm not really sure ...
2
votes
1answer
95 views

Prove the inequality $\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy} \geq \frac{3}{1+\frac{(x+y)^2}{4}}$ when $x^2+y^2=1$

I have to prove the inequality $$ \frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy} \geq \frac{3}{1+\frac{(x+y)^2}{4}} $$ when $x^2+y^2=1$, using Cauchy-Schwarz Inequality. The RHS is equal to $\frac{12}...
1
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2answers
69 views

Find the minimum and maximum values of $P=\left( 6-a^2-b^2-c^2\right)\left(2-abc\right)$

Let $a,b,c$ be non-negative real numbers such that $c \geq 1$ and that $a+b+c=2$. Find the minimum and maximum values of $$P=\left( 6-a^2-b^2-c^2\right)\left(2-abc\right)$$ To find the minimum of $P$ ...
1
vote
4answers
77 views

Proving $\frac{1}{6}a+\frac{1}{3}b+\frac{1}{2}c \geq \frac{6abc}{3ab+bc+2ca}$ for positive $a$, $b$, $c$

I'm at the end of an inequality proof that started out complex and I was able to simplify it to: $$\frac{1}{6}a+\frac{1}{3}b+\frac{1}{2}c \geq \frac{6abc}{3ab+bc+2ca} \quad\text{where}\quad a, b, c ...
0
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0answers
25 views

A tough inequality with a constraint?

Let $x_1,$ $x_2,$ $x_3,$ $x_4,$ $x_5$ be distinct positive integers such that $x_1 + x_2 + x_3 + x_4 + x_5 = 100.$ Compute the maximum value of the expression \begin{align*} &\frac{(x_2 x_5 + 1)(...
0
votes
1answer
33 views

Does the opposite inequality of A.M ≥ G.M holds?

Let $a,b>0$. Does there exists $C_1,C_2>0$ such that $a+b \leq C_1(ab)^{C_2}$ holds? We know from A.M ≥ G.M that the opposite inequality holds for $C_1=2,C_2=½$
1
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2answers
49 views

Area of a rectangle inside a triangle with given coordinates

Given a triangle with vertices at points $(0, -a), (0, a), (b, 0)$, where $a > 0$, find the maximal area and the dimensions (base and height) of a rectangle that can be contained within the ...
5
votes
2answers
68 views

Inequality related with $abcd=(1-a)(1-b)(1-c)(1-d)$ [duplicate]

Given that $0<a,b,c,d<1$ satisfying $abcd=(1-a)(1-b)(1-c)(1-d)$. Prove that $$(a+b+c+d)-(a+c)(b+d)\geq 1.$$ First, I have already done a quite similar exercise as below: "Given that $a^2+b^2+c^...
2
votes
1answer
93 views

prove this inequality with $x^3+y^3+z^3$

Let $x,y,z>0$. Show that $$x^3+y^3+z^3+2(xy^2+yz^2+zx^2)\ge 3(x^2y+y^2z+z^2x).$$ I have a solution by using $y=x+a,z=x+a+b$; my question is: can this be solved using simple methods (such as AM-...
2
votes
2answers
67 views

How to correctly find the value of theta for which $\frac{a}{\cos\theta}+\frac{b}{\sin\theta}$ is minimum?

I was solving a question which required me to find the value of $\theta$ for which the expression $\frac{a}{\cos\theta}+\frac{b}{\sin\theta}$ has its minimum value. Given that, $a=3\sqrt3, b=1$. ...
1
vote
3answers
49 views

Upper bound estimate $ab^3\leq c(a^2+b^2)^2$. Find optimal, or good, constants $c$

I am trying to find a good upper bound estimate for the expression $ab^3$, where $a,b\in\mathbb R$, and it should be of the form $ab^3\leq c (a^2+b^2)^2, c\in\mathbb R.$ (The reason for the latter is,...
2
votes
2answers
221 views

show this inequality to $\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $

Let $a,b$ and $c$ be positive real numbers. Prove that $$\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $$ This problem is from Iran 3rd round-2017-Algebra final exam-P3,...
3
votes
4answers
114 views

What is wrong with this solution of find the least value of $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$

Find the least value of $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$ They all are positive terms so arithmetic mean is greater than equal to geometric mean $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x\geq 3( ...
0
votes
2answers
72 views

Prove that $\frac{2-2a^2}{1+a^2}+\frac{1-b^2}{1+b^2}+\frac{1-c^2}{1+c^2} \leq \frac{9}{4}$

Let $a,b,c$ be positive real numbers such that $ab+bc+ca=1$. Prove that $$\frac{2-2a^2}{1+a^2}+\frac{1-b^2}{1+b^2}+\frac{1-c^2}{1+c^2} \leq \frac{9}{4}$$ I find when the equality occurs at $a=\frac{1}{...
2
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1answer
33 views

AM-GM inequality for functions

As all you masters in math know, the famous AM-GM inequality in positive numbers is as follows: $${1\over n}\sum_{k=1}^{n}x_k\ge \left(\prod_{k=1}^{n}x_k\right)^{1\over n}$$ which can be rewritten ...
4
votes
3answers
151 views

Prove that $\frac{1}{ab}+\frac{1}{cd} \geq \frac{a^2+b^2+c^2+d^2}{2}$

Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that $$\frac{1}{ab}+\frac{1}{cd} \geq \frac{a^2+b^2+c^2+d^2}{2}$$ I write $$a^2+b^2+c^2+d^2=16-2\left(ab+cd+\left(a+b\right)\left(c+...
1
vote
3answers
70 views

About the fact that $\dfrac{a^2}{b} + \dfrac{b^2}{a} + 7(a + b) \ge 8\sqrt{2(a^2 + b^2)}$.

There's a math competition I participated yesterday (19/3/2019). In these kinds of competitions, there will always be at least one problem about inequalities. Now this year's problem about ...
1
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2answers
117 views

Difficult inequality with 4 variables

I am struggling with this inequality from the book Advanced Olympiad Inequalities: Algebraic & Geometric Olympiad Inequalities, any idea please? Thanks. Question: Let $a,b,c,d>0$ such that $a^...
1
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1answer
64 views

Given $a+b+c+d=4$, find the minimum value of $\Sigma_{cyc}\frac{a}{b^3+4}$

Given $a+b+c+d=4$, find the minimum value of $\Sigma_{cyc}\frac{a}{b^3+4}$ I'm pretty sure this has something to do with Holder's inequality, but I don't know how to solve this. By guessing I found $a=...
1
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2answers
86 views

Prove that $a\sqrt{b^3+1}+b\sqrt{c^3+1}+c\sqrt{a^3+1} \leq 5$

Let $a,b,c$ be nonnegative real numbers such that $a+b+c=3$. Prove that $$a\sqrt{b^3+1}+b\sqrt{c^3+1}+c\sqrt{a^3+1} \leq 5$$ I found a point at which the equality is attended, say $a=0,b=1,c=2$. But I ...
0
votes
2answers
63 views

Find the minimum value of $abc$.

$a$, $b$, $c$ are three positives and $m$, $n$, $p$, $x$, $y$, $z$ are positive parameters . Find the minimum value of $abc$ such that the following inequation is correct. $$\large \dfrac{x}{x + ma} + ...
0
votes
6answers
85 views

Solve the equation $x^2 + 4(\sqrt{1 - x} + \sqrt{1 + x}) - 8 = 0$

Solve the equation $x^2 + 4(\sqrt{1 + x} + \sqrt{1 - x}) - 8 = 0$. Let $\sqrt{1 + x} = a$, $\sqrt{1 - x} = b$. I tried doing this. "$1 - x^2 = [\sqrt{(1 - x)(1 + x)}]^2 = (ab)^2$. The original ...
0
votes
4answers
185 views

Find the maximum of the expression: $ \frac{(а-b)^2(b-с)^2(с-а)^2}{ ((а-b)^2+(b-с)^2+(с-а)^2)^3}$

Find the maximum of the expression : $$ \frac{(а-b)^2(b-с)^2(с-а)^2} { ((а-b)^2+(b-с)^2+(с-а)^2)^3}$$ At first I tried to solve this problem with the help of inequality A.M-G.M, I know the answer ...
2
votes
1answer
100 views

Prove the inequality $\sum x+6\ge 2(\sum\sqrt{xy}) $

Let $x;y;z\in R^+$ such that $x+y+z+2=xyz$. Prove that $$x+y+z+6\ge 2(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}) $$ This inequality is not homogeneous and look at the condition i thought that i would ...
0
votes
0answers
44 views

prove the following inequality using am gm hm or weirstrass etc

For $a_0,a_1,a_2,......,a_n \in R, \,\, a_0<a_1<a_2<....<a_n$ show that $$ \frac n{a_1-a_0}+\frac {n-1}{a_2-a_1}+....+\frac 1{a_n-a_{n-1}} \ge \sum_{k=1}^n \frac {k^2}{a_k} $$ i recently ...
-1
votes
1answer
29 views

prove the inequality using inequalities like AM GM HM OR CAUCHY or WEIRSTRASS ETC.

The inequality to be proven is $$ 2^n \gt 1 + n\cdot \sqrt{2^{n-1}} for\ all\ n>2 $$ using any inequalities like am gm hm cauchy schwarz tchebychev etc I recently studied inequalities came across ...
1
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2answers
56 views

Akerberg's Refinement of AM-GM: Motivation of proof

In Steele's Cauchy Schwarz Master Class Exercise 2.10 is about Akerberg's Refinement of AM-GM. (It's a refinement because AM-GM follows by iteration.) Problem. For $a_1, \dots, a_n \geq 0$ and $n \...
2
votes
1answer
71 views

$f\left ( a,\,b,\,c \right )\leqq f\left ( a,\,1,\,c \right )$ with $abc= 1$ and $a,\,b,\,c> 0$

Give $3$ positve numbers $a,\,b,\,c$ such that $abc= 1$ , prove: $$f\left ( a,\,b,\,c \right )= \frac{3\,a+ 2\,b}{\sqrt{5\,a^{\,2}- ab+ b^{\,2}}}+ \frac{3\,b+ 2\,c}{\sqrt{5\,b^{\,2}- bc+ c^{\,2}}}\...
0
votes
4answers
63 views

Some Cauchy-Schwarz Inequalities [closed]

I am trying to learn how to deal with inequalities to prepare for a Math Olympiad and right now I am working on Cauchy-Schwarz. However, I am not that good at seeing the relationships and I don't have ...
0
votes
1answer
39 views

Minimum distance of the curve $xyz^2=2$ from its origin

Consider a $3D$ figure represented by $xyz^2=2$. Then what is its minimum distance from its origin? What I try: Given $xyz^2=2$ and I have to find minimum of $x^2+y^2+z^2$ Let $x^2+y^2+z^2=k^2$ ...
3
votes
1answer
108 views

Inequality proof (strange)

Given $a^2 +b^2 +c^2 +d^2 =1$ where $a,b,c,d$ are positive real numbers, prove that $a+b+c+d-1 \geq 16abcd$ How can I prove the inequality ? My attempts: By Cauchy-Schwarz : $ (a+b+c+d)^{2} \leq (...
2
votes
2answers
119 views

How to prove the follwing inequality $(a+b)(b+c)(c+a) \ge 8$

If $$a+b+c=3abc$$ and $$a,b,c > 0$$ prove that $$(a+b)(b+c)(c+a)\geq 8$$ I can fairly easily prove that $(a+b)(b+c)(c+a)\geq8abc$, but then I get stuck.....since then I cannot move forward ...
0
votes
1answer
32 views

Maximise right circular cone volume with fixed surface area using inequalites

This is a question inspired from What is the relation between height and radius of base of a right circular cone when its surface area is given and volume is to be maximum? In the linked post, OP has ...
3
votes
1answer
129 views

Find $\max k$ in the inequality $y^2 - xy + x + 2y + 5 \geq k\,(\sqrt{3x} + \sqrt{y})$ for $0\le x,y\le 3$.

The inequality $$y^2 - xy + x + 2y + 5 \geq k\,(\sqrt{3x} + \sqrt{y})$$ holds for all $x$ and $y$ such that $0\leq x,y\leq 3$. Find maximum $k$. The answer is $2$. Can someone please solve it? ...
4
votes
3answers
84 views

What's maximum value of $x (1-x^2)$ for $0 < x <1$?

Since we are being taught about AM-GM inequality, I decided to use the method however I am getting two different answers by slightly different methods. Method 1 \begin{equation} v=x (1-x^2)$ \implies ...
-1
votes
1answer
68 views

If $x,y,z \in \mathbb{R}^+$, then prove that $\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x}\geq x+y+z$ [closed]

I think that it can be proved by the arthemetic and geometric means in the relation comes $$ \frac{x+y+z}{3} \geq (xyz)^{\frac{1}{3}}$$
1
vote
1answer
255 views

Stronger than AM-GM and a conjecture

Let $a_i>0$ be $n$ numbers such that $\prod_{i=1}^{n}a_i\leq 1$ then we have: $$\sum_{i=1}^{n}a_i\geq n \left(\prod_{i=1}^{n}a_i \right)^{\large\left({n+\sum_{i=1}^{n}a_i-n(\prod_{i=1}^{n}a_i)^{1/...
0
votes
2answers
34 views

Proving an inequality using am-gm inequality

Prove that if $x\neq y\neq z$, $x,y,z\gt0$ $$\frac{x^2y^2+y^2z^2+z^2y^2}{x+y+z}\geqslant xyz$$ I tried using $$x^2y^2+y^2z^2+z^2x^2\gt 3\root3\of{x^4y^4z^4}$$ $$x+y+z\gt 3\root3\of{xyz}$$ But I can't ...
2
votes
2answers
66 views

Find maximum value by using AM-GM inequality

I have a problem: Find the maximum value of $P=\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}+\frac{y^3z^4x^3}{(y^4+z^4)(yz+x^2)^3}+\frac{z^3x^4y^3}{(z^4+x^4)(zx+y^2)^3}$ with $x,y,z>0$. Is there anyway to ...
1
vote
1answer
41 views

For three positive numbers verifying $xyz=2+x+y+z$, is there an upper bound on $x+y+z$? And, if so, which is that?

For three positive numbers verifying $xyz=2+x+y+z$, is there an upper bound on $x+y+z$? And, if so, which is that? I've managed to find only a lower bound, from $ xyz=2+x+y+z \le \frac{(x+y+z)^3}{27}...
3
votes
2answers
77 views

Understanding a proof from the APMO 1998 on inequalities.

I was having trouble with proving the following inequality.The question was from the book Secrets to Inequalities by Pham Kim Hung. $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} \geq \frac{x+y+z}{\sqrt[3]{...
1
vote
3answers
213 views

Prove that $a^2+b^2+c^2\geqslant\frac{1}{3}$ given that $a\gt0, b\gt0, c\gt0$ and $a+b+c=1$, using existing AM GM inequality [closed]

Using the AM and GM inequality, given that $a\gt0, b\gt0, c\gt0$ and $a+b+c=1$ prove that $$a^2+b^2+c^2\geqslant\frac{1}{3}$$
2
votes
4answers
189 views

Proving an equality using given ones; no need for differentiation

Prove that $(\frac ab+\frac bc+\frac ca)(\frac ba+\frac cb+\frac ac)\geqslant9$ The formulas given were $$\frac{a+b} {2}\geqslant\sqrt {ab}$$ $$a^2+b^2\geqslant2ab$$ $$\frac{a+b+c} {3}\geqslant\root3\...
2
votes
2answers
84 views

In any triangle is $\sin A+\sin B+\sin C=\frac{3\sqrt3}{2}$ always

Well I came with an interesting proof. But I just want to verify it From here we will get $$\sin A+\sin B+\sin C\leq \frac{3\sqrt3}{2}.$$ and from this I get $$\sin A+\sin B+\sin C\geq \frac{3\sqrt3}...
4
votes
6answers
145 views

prove that : $\dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} \geq abc$ for $a ,b ,c \in \mathbb{R}^{>0}$

prove that : $\dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} \geq abc$ for $a ,b ,c \in \mathbb{R}^{>0}$ . I think that must I use from $\dfrac{a^2}{2}+\dfrac{b^2}{2} \geq ab$ but no result please ...