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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.

0
votes
2answers
22 views

Inequality problem (may be) involving means

If n is a positive integer then how can I prove that $$2^n>1+n \sqrt{2^{n-1}}$$ .Any hint may help.My textbook mentions this problem in category of A.M. ,G.M. , H.M. inequalities.So please give ...
0
votes
1answer
33 views

How can we apply AM - GM on two variables?

Let a and b be the positive numbers. Find the least value of a + (a/b) + (1/a) + b. I applied AM -GM inequality on five numbers a, a/b, 1/(2a), 1/(2a) and b to get the answer as 5/(4^(1/5)). However,...
0
votes
1answer
38 views

Inequality involving a kind of Harmonic mean

While revising the Harmonic mean, I came across this inequality which I haven't figured out how to solve, but I think it should be the application of some known inequality. I would be very grateful if ...
6
votes
2answers
90 views

Prove following inequality

Prove that $(\frac{2a}{b+c})^\frac{2}{3}+(\frac{2b}{a+c})^\frac{2}{3}+(\frac{2c}{a+b})^\frac{2}{3} ≥ 3$ What I tried was to use AM-GM for the left side of this inequality, what I got was $3(\frac{8abc}...
1
vote
2answers
60 views

Proving the given inequalities

Q: Prove the given inequalities for positive a,b,c:$(i) \left[\frac{bc+ca+ab}{a+b+c}\right]^{a+b+c}>\sqrt{(bc)^a.(ca)^b.(ab)^c}$$(ii) \left(\frac{a+b+c}{3} \right)^{a+b+c}<a^ab^bc^c<\left(\...
0
votes
4answers
40 views

Infimum and supremum of $x^5y^2z$

$A=\{x^5y^2z : x,y,z>0$ and $x+y+z=7\}$ In my mind $\sup A=5^5$ and $\inf A=0$, but how can I prove this?
0
votes
2answers
36 views

Max value of sum of squares

Question is Sum of two numbers x and y is 10. What is the maximum and minimum value of its sum of squares. How can I find the maximum? The minimum turns out to be 50 using AM GM Please do try to ...
2
votes
4answers
84 views

Let $a, b, c \in \mathbb{R^+}$ and $abc=8$ Prove that $\frac {ab+4}{a+2} + \frac {bc+4}{b+2} + \frac {ca+4}{c+2} \ge 6$

Let $a, b, c \in \mathbb{R^+}$ and $abc=8$ Prove that $$\frac {ab+4}{a+2} + \frac {bc+4}{b+2} + \frac {ca+4}{c+2} \ge 6$$ I have attempted multiple times in this question and the only method that I ...
4
votes
3answers
91 views

Maximise $(x+1)\sqrt{1-x^2}$ without calculus

Problem Maximise $f:[-1,1]\rightarrow \mathbb{R}$, with $f(x)=(1+x)\sqrt{1-x^2}$ With calculus, this problem would be easily solved by setting $f'(x)=0$ and obtaining $x=\frac{1}{2}$, then checking ...
-1
votes
5answers
77 views

Inequality for contests [closed]

Prove that for real numbers $x,y,z\in[0;1/2]$ with $x+y+z=1 :$ $$ \sqrt{1-x^2} + \sqrt{1-y^2} + \sqrt{1-z^2} \geq 2\sqrt{2}$$
3
votes
2answers
86 views

Maximize the value of $\sqrt{x-x^2}+\sqrt{cx-x^2}$ without using calculus

Assume that $c$ is positive. How can we maximize the value of $\sqrt{x-x^2}+\sqrt{cx-x^2}$ with respect to $x$ without the use of calculus? With calculus, we can easily find out that the max of the ...
1
vote
2answers
82 views

Proving $\frac{3}{a^{2}b^{2}c^{2}}+\sum{\frac{a^{11}}{bc}}\geq \frac{\sum{a^{6}}+9}{2}$ [closed]

For all positive integers $a, b, c$, prove that: $$\frac{3}{a^{2}b^{2}c^{2}}+\sum{\frac{a^{11}}{bc}}\geq \frac{\sum{a^{6}}+9}{2}$$ And $\sum$ denotes cyclic sum. My attempt was writing the LHS as $\...
-1
votes
1answer
128 views

Prove that $ \sqrt{xy} + \sqrt{xz} + \sqrt{xt} + \sqrt{yz} + \sqrt{yt} + \sqrt{zt} \geq 3{( xyz + xyt + xzt + yzt)}^{\frac{1}{3}}$ [closed]

Let $ x, y, z, ,t$ four real positive number. Prove that $$ \sqrt{xy} + \sqrt{xz} + \sqrt{xt} + \sqrt{yz} + \sqrt{yt} + \sqrt{zt} \geq 3{( xyz + xyt + xzt + yzt)}^{\frac{1}{3}}$$ I tried using ...
0
votes
2answers
49 views

problem regarding application of Jensen's inequality

question: For $a,b,c,d \in \mathbb{R^+}$ with $a+b+c+d = 4$, Prove $\displaystyle \sum\dfrac{a}{b(b+1)}\geq \dfrac{8}{(a+c)(b+d)}$ my attempt: $f(x)= \dfrac{1}{x^2+x}=\dfrac{1}{x(x+1)}$ is convex ...
2
votes
2answers
103 views

Find the minimum values of $a^3$ and $b^3$ from the following given cubic function.

Today after teaching AM,GM,HM relations my teacher proposed this question. It isn't a HW though, since we don't have to tell that we solved the question and submit a solution. Question Given ...
3
votes
3answers
100 views

Prove $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}} \le \frac{3}{2}$

If $a.b,c \in \mathbb{R^+}$ and $ab+bc+ca=1$ Then Prove $$S=\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}} \le \frac{3}{2}$$ My try we have $$S=\sum \frac{a}{\sqrt{a^2+ab+bc+...
0
votes
2answers
53 views

$(1-x)^{3}|1-xe^{iy}|^{4}|1-xe^{2iy}|^{2}<1$ for real $x,y$ with $x\in(0,1)$

For real $x,y$ with $x\in(0,1)$ we have $$(1-x)^{3}|1-xe^{iy}|^{4}|1-xe^{2iy}|^{2}<1$$ My attempt : Obviously, we have the identity $(1-x)^{3}|1-xe^{iy}|^{4}|1-xe^{2iy}|^{2}=(1-x^{3})|1-x\cos y-...
0
votes
1answer
60 views

How to prove the given inequality using CS or AM GM HM inequality [closed]

I was solving a problem on triangular inequalities and I have ended up with the following inequality required to be proven $$\sum_{cyc}\frac{x}{2x+y+z}\geq\frac{9\sqrt{3(x+y+z)xyz}}{4(x+y+z)^2},$$ ...
0
votes
1answer
30 views

Two versions of AM-GM inequality

I'm currently studying functional analysis and in the preliminary chapter the author gives the following inequality: AM-GM Inequality: Let $x,y>0$ and $0< \lambda <1$. Then $$x^\...
0
votes
0answers
47 views

How to manipulate the following inequality [duplicate]

there was this solved question i was doing $a,b,c$ are positive real numbers with sum $3$ . prove that $\sqrt{a}+\sqrt{b}+\sqrt{c} \geq ab +bc+ca $ in the solution the author ,in first step ,said ...
0
votes
1answer
16 views

Why do we usually require $b<a$ in A.M.-G.M. Inequality?

The A.M.-G.M. is sometimes stated as follows. Let $a$ and $b$ are positive numbers such that $b < a$. Let $a_1 = a$ and $b_1 = b$. Define the sequences $\{a_n\}$ and $\{b_n\}$ recursively for all ...
5
votes
3answers
192 views

For $a\geq2$, $b\geq2$ and $c\geq2$, prove that $\left(a^3+b\right)\left(b^3+c\right)\left(c^3+a\right)\geq125 abc$

For $a\geq2$, $b\geq2$ and $c\geq2$, prove that $$(a^3+b)(b^3+c)(c^3+a)\geq 125 abc.$$ My try: First I wrote the inequality as $$\left(a^2+\frac{b}{a}\right) \left(b^2+\frac{c}{b}\right) \left(...
0
votes
2answers
142 views

A very strange algebra problem with four numbers with an average 1.

Here is a Russian algebra problem from a 1999 olympiad. I can’t solve it, please help! Find all possible values of $a,b,c,d$, if they are positive reals, their average is 1, and $$\dfrac{3-a+b(-a-ac)}...
10
votes
6answers
261 views

How to compare logarithms $\log_4 5$ and $\log_5 6$?

I need to compare $\log_4 5$ and $\log_5 6$. I can estimate both numbers like $1.16$ and $1.11$. Then I took smallest fraction $\frac{8}{7}$ which is greater than $1.11$ and smaller than $1.16$ and ...
2
votes
1answer
87 views

How to prove inequality $(a^2+b^2+c^2)^3\ge6(a^3+b^3+c^3)^2$ when $a+b+c=0$?

I know the identity $a^3+b^3+c^3-3abc = 1/2 (a+b+c) [(a-b)^2+(b-c)^2+(c-a)^2]$. So when it comes to this problem where $a+b+c=0$, you get $(a^2+b^2+c^2)^3\ge54(abc)^2$ when $a+b+c=0$ (because $ a^3+b^...
7
votes
1answer
154 views

For positve $a$, $b$, $c$, $d$ with $a+b+c+d\leq 1$, prove that $\sqrt[4]{(1-a^4)(1-b^4)(1-c^4)(1-d^4)}\geq255\cdot a b c d .$

Let $a,b,c,d\in\mathbb R_+$ such that $a+b+c+d\leqslant1$. Prove that$$ \sqrt[4]{\smash[b]{(1-a^4)(1-b^4)(1-c^4)(1-d^4)}}\geqslant255·abcd. $$ My observations: I can see that all of $a,b,c,d$ are ...
0
votes
1answer
115 views

Prove the inequality $\sqrt{\frac{a^2+1}{b+c}}+\sqrt{\frac{b^2+1}{a+c}}+\sqrt{\frac{c^2+1}{a+b}}\ge 3$

Let $a;b;c\in R^+$ such that $ab+bc+ca>0$. Prove that $$\sqrt{\frac{a^2+1}{b+c}}+\sqrt{\frac{b^2+1}{a+c}}+\sqrt{\frac{c^2+1}{a+b}}\ge 3$$ I have seen the similar question is $$\frac{a^2+1}{b+c}+\...
7
votes
1answer
54 views

Inequality, how to know intuition behind it

I was solving the following inequality For $a$, $b$, $c$ and $d$ being positive real numbers which goes as $$ \frac{a}{a+b} + \frac{b}{b+c} + \frac{c}{c+d} + \frac{d}{d+a} \leq \frac{a}{b+c} + \frac{...
4
votes
2answers
131 views

P.T. $\frac{1}{a^3(b+c)} +\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)} \ge \frac 32$

If $abc=1$ where $a,b,c$ are positive real. Prove that ,$\frac{1}{a^3(b+c)} +\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)} \ge \frac 32$. I tried to multiply the LHS by $abc$ to make the relation homogeneous ...
0
votes
0answers
36 views

How to prove the following inequality for a function?

I have function $$D(M\gamma)=\frac{M\gamma K}{K+\gamma(M\gamma-1)}\tag{1}$$ where $K$ is some positive constant. In this case, how to show that $$D(M(\gamma-1))<D(M\gamma),~~\text{for } \gamma<\...
1
vote
2answers
104 views

How to prove this inequality using AM-GM?

Suppose $a,b,c$ are positive real numbers. Then prove that $$\Big(\frac{a+b}{2}\Big)\Big(\frac{b+c}{2}\Big)\Big(\frac{c+a}{2}\Big)\ge\Big(\frac{a+b+c}{3}\Big)\Big(abc\Big)^\frac{2}{3}\tag{*}$$ My ...
1
vote
1answer
39 views

Show that for positive numbers a,b,c,d, $\sum_{cyc} ab \leq \frac{1}{4}\left(\sum_{cyc} a \right)^2$ and … [duplicate]

Let a,b,c,d be four positive real numbers. Show that $$\sum_{cyc} ab \leq \frac{1}{4}\left(\sum_{cyc} a \right)^2$$ and $$\sum_{cyc} abc \leq \frac{1}{16}\left(\sum_{cyc} a \right)^3$$ My textbook ...
0
votes
2answers
85 views

Maximizing $x^2 y + y^2 z + z^2 x − x^2 z − y^2 x − z^2 y$ for $x$, $y$, $z$ between $0$ and $1$ (inclusive)

Find the maximum value of the expression:- $$x^2 y + y^2 z + z^2 x − x^2 z − y^2 x − z^2 y$$ when $0 \leq x \leq 1$, and $0 \leq y \leq 1$, and $0 \leq z \leq 1$. Please note: This is a initial ...
9
votes
3answers
172 views

Prove that if $a+b+c+d=4$, then $(a^2+3)(b^2+3)(c^2+3)(d^2+3)\geq256$

Given $a,b,c,d$ such that $a + b + c + d = 4$ show that $$(a^2 + 3)(b^2 + 3)(c^2 + 3)(d^2 + 3) \geq 256$$ What I have tried so far is using CBS: $(a^2 + 3)(b^2 + 3) \geq (a\sqrt{3} + b\sqrt{3})^2 = ...
2
votes
2answers
67 views

Prove this inequality for given conditions

For all $x,y>0$, $$\frac{1}{(x+1)^2} + \frac{1}{(y+1)^2} \ge \frac{1}{xy+1}$$ I can only think of substituting $x+1$ with $a$ and $y+1$ with $b$. Then the inequality turns into $$(a^2 + b^2) (ab-a-...
0
votes
4answers
103 views

How to prove $(xyz)^{1/3} \le (x+y+z)/3$ using linear algebra

To elaborate on the title, you can prove that $$\sqrt{xy}\le \frac{x+y}{2}$$ in the following way: Is there a way that this can be extended to the inequality in the title and to a general case? For ...
2
votes
2answers
99 views

find the range $(x_{1}+x_{2}+\cdots+x_{2009})^2=4(x_{1}x_{2}+x_{2}x_{3}+\cdots+x_{2009}x_{1})^2$

let $x_{i}\ge 0(i=1,2,3,\cdots,2009$,and such $$(x_{1}+x_{2}+\cdots+x_{2009})^2=4(x_{1}x_{2}+x_{2}x_{3}+\cdots+x_{2009}x_{1})=4$$ find the range $\sum_{i=1}^{2009}x^2_{i}$ I guess the range is $[1.5,...
4
votes
4answers
99 views

How to show that: $\left(\frac{mn+1}{m+1}\right)^{m+1} > n^m$

If m and n are positive integers then show that:$$\left(\frac{mn+1}{m+1}\right)^{m+1} > n^m$$I am new in this Course.So i can't able to think how i start a inequalities question by looking it's ...
1
vote
1answer
157 views

If $x+y+z=3$ then $\sum x\sqrt{x^3+3y} \ge 6$

Let $x,y,z>0$ such that $x+y+z=3$. Prove that $$\sum x\sqrt{x^3+3y} \ge 6$$ This trying doesn't help. With Cauchy Schwarz $(\sum x\sqrt{x^3+3y})^2\geq \sum x^2\sum(x^3+3y) = (x^2 + y^2 + z^2)(x^...
0
votes
2answers
108 views

$X>0$, $Y>0$ and $X^2+Y ^3\geq X^3+Y^4$. Prove that $X^3+Y ^3\leq 2$ [duplicate]

$X>0$, $Y>0$ and $X^2+Y ^3\geq X^3+Y^4$. Prove that $X^3+Y ^3\leq 2$ First I tried: $0<X\leq1$ and $0<Y\leq1$, but this in not the only case for $X^2+Y ^3\geq X^3+Y^4$ and get nowhere. ...
-3
votes
2answers
300 views

Inequality $\sqrt{xy+yz+zx} \ge \frac {8}{15} (x+y+z)$ [closed]

By Titu's inequality: $\sum_{cyc} \frac {1}{x+y} \ge \frac {(1+1+1)^2}{2(x+y+z)} = \frac {9}{2(x+y+z)}$ Then, to prove: $ \frac {3}{x+y+z} + \sum_{cyc} \frac {1}{x+y} \ge \frac {4}{\sqrt{xy+yz+zx}}$...
0
votes
4answers
78 views

Prove that $a_n \gt b_n$ $\forall $ $n \ge 6$

Given that $$a_n =\left(1^2+2^2+3^2+\cdots+n^2\right)^n$$ and $$b_n=n^n \times (n!)^2$$ Then prove that $a_n \gt b_n$ $\forall $ $n \ge 6$ My attempt: I tried using induction, but I could not ...
1
vote
1answer
66 views

Generalized Sum over Product Inequality : Related to Cauchy Schwarz's and AM-GM Inequality.

Consider set $S = \left \{ a_i \mid 1 \leq i \leq n \right \}$ where $a_i \in \mathbb{R}$ and fixed integer $t$ I do not know whether the following inequality is true or not: $$\sum_{P \subset S, \...
2
votes
4answers
105 views

How to prove AM-GM by induction 3

Let $a_1;a_2;...;a_n\ge 0$. Prove that $$\frac{\sum ^n_{k=1}a_k}{n}\ge \sqrt[n]{\prod ^n_{k=1}a_k}$$ We will prove it's true with $n=k$. Indeed we need to prove it's true with $n=k+1$ WLOG $a_1\le ...
2
votes
5answers
69 views

Prove that $\log_n(n+1)\geq\log_{n+1}(n+2)$ for $n>1$

Prove that $\log_n(n+1)\geq\log_{n+1}(n+2)$ for $n>1$. So far I only know that $\log_n(n+1)>\frac{\log_n(n+2)}{\log_n(n+1)}$ Since $n>1$, LHS must be greater than RHS. Is there any other ...
7
votes
2answers
925 views

Largest possible value of trigonometric functions

Find the largest possible value of $$\sin(a_1)\cos(a_2) + \sin(a_2)\cos(a_3) + \cdots + \sin(a_{2014})\cos(a_1)$$ Since the range of the $\sin$ and $\cos$ function is between $1$ and $-1$, shouldn'...
2
votes
1answer
144 views

Prove that: $\frac{bc}{a^2+1}+\frac{ac}{b^2+1}+\frac{ab}{c^2+1}\leq \frac{3}{4}$ [duplicate]

Given three positive numbers a,b,c satisfying $$a^2+b^2+c^2=1$$ Prove that: $$\frac{bc}{a^2+1}+\frac{ac}{b^2+1}+\frac{ab}{c^2+1}\leq \frac{3}{4}$$ The things I have done so far: $$\sum \limits_{cyc}\...
1
vote
2answers
260 views

Find the maximum of the value$(n^{n-1}-1)\sqrt[n]{a_{1}a_{2}\cdots a_{n}}+\sqrt[n]{\frac{a^n_{1}+a^n_{2}+\cdots+a^n_{n}}{n}}$

Let $n$ be give a positive integer, $a_{i} \ge 0$,such that $a_{1}+a_{2}+\cdots+a_{n}=n$. Find the maximum value of $$(n^{n-1}-1)\sqrt[n]{a_{1}a_{2}\cdots a_{n}}+\sqrt[n]{\dfrac{a^n_{1}+a^n_{2}+\...
2
votes
2answers
166 views

Prove that: $xy\sqrt{z}+yz\sqrt{x}+zx\sqrt{y}\geq x+y+z$

Let $x$,$y$ and $z$ are positive and $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\leq 3$$ Prove that: $$xy\sqrt{z}+yz\sqrt{x}+zx\sqrt{y}\geq x+y+z$$ The things I have done so far $$3\geq \sum \limits_{cyc}\...