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.

The AM-GM states that, given a finite number of non-negative real numbers, their arithmetic mean is always greater than or equal to their geometric mean (hence the name “AM-GM inequality”) and that the equality holds if and only if all the given numbers are equal. In other words, if $a_1,a_2,\ldots,a_n\in(0,+\infty)$, then$$\frac{a_1+a_2+\cdots+a_n}n\geqslant\sqrt[n]{a_1a_2\ldots a_n}$$and$$\frac{a_1+a_2+\cdots+a_n}n=\sqrt[n]{a_1a_2\ldots a_n}\iff a_1=a_2=\cdots=a_n.$$

A weighted version of AM-GM inequality is $$\frac{w_1a_1+w_2a_2+\cdots+w_na_n}w\geqslant\sqrt[w]{a_1^{w_1}a_2^{w_2}\ldots a_n^{w_n}}$$

where $w_i$ are nonnegative and $w=\sum_{i=1}^nw_i$.

1450 questions
8
votes
1 answer

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 $$ \frac{a}{a+b} + \frac{b}{b+c} + \frac{c}{c+d} + \frac{d}{d+a} \leq \frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b} $$ Which…
pde
  • 790
3
votes
3 answers

How to check that the arithmetic mean of any subset $S\ne\{1\}$ of $A={1,2,...,n}$ is at least $\frac 32$?

How to check that the arithmetic mean of any subset $S\ne\{1\}$ of $A=\{1,2,...,n\}$ is at least $\frac 32$? I don't have any idea where to start this. It is clear that this is right and I need to demonstrate because it would help me solve another…
IONELA BUCIU
  • 1,157
  • 1
  • 17
3
votes
0 answers

How to prove $\dfrac{a+b}{2}\geq\sqrt{ab}$ using ellipse

Noted that ellipse properties $d_1+d_2=2D$, focal length, $f=c$ and radius of minor axis $=r$. Let $d_1=a;d_2=b$ If $ab$ is not maximum, then $\sqrt{ab}$ not maximum W.l.o.g, prove max…
Pck Tsp
  • 135
3
votes
2 answers

Using AM-GM inequality prove that $(1+\sqrt{2})(1+\sqrt{3})(1+\sqrt{5}) \lt 8\sqrt{30}$.

It is trivial to prove that $(1+\sqrt{2})(1+\sqrt{3})(1+\sqrt{5}) \le 8\sqrt{30}$ using numeric methods. For example by multiplying $(1+\sqrt{2}) \le 3 $ $(1+\sqrt{3}) \le 3 $ $(1+\sqrt{5}) \le 4 $ We get: $(1+\sqrt{2})(1+\sqrt{3})(1+\sqrt{5})…
motoras
  • 733
2
votes
0 answers

$ \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c}=2 $ then $ \sum_{cyc}\sqrt{ab} \geq \frac{3}{2}$

Prove that if $$ \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c}=2 $$ then $$\sum_{cyc} \sqrt{ab} \geq \frac{3}{2}$$ I tried to use AM-GM inequality and Titu's lemma and was able to show that $a+b+c\geq3$ and $\sum_{cyc}\sqrt{ab}\geq3abc$. Can…
2
votes
0 answers

Basic Inequality problem related to AM-GM Inequalities

I am looking to prove $$ \frac{ab}{a⁵+b⁵+ab} + \frac{bc}{b⁵+c⁵+bc} + \frac{ac}{a⁵+c⁵+ac} \le1 $$ Where $ a,b,c $ are positive reals with $ abc = 1 $ My initial thought was to apply AM-GM to get $$ a⁵ + b⁵ + ab \ge3(a²b²) $$ and substituting this in…
2
votes
1 answer

AM GM inequality using a relation

I am given a relation that for $a,b>0$ we have $$ab\leq \dfrac{a^p}{p}+\dfrac{b^q}{q}$$ where $ p$ and $ q$ satisfy $\dfrac{1}{p}+\dfrac{1}{q}=1$. Now how do I use this to prove the AM-GM inequality?
Upstart
  • 2,613
2
votes
4 answers

For $x\geq 0$, what is the smallest value of $\frac{4x^2+8x+13}{6(x+1)}$?

I know that I have to use the AM-GM inequality. I tried separating the fraction: $$\frac{4x^2+2x+7}{6(x+1)} + \frac{6(x+1)}{6(x+1)}$$ However, it doesn't seem to make either side of the inequality into a number. I would appreciate some help,…
Ayo
  • 111
1
vote
2 answers

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 minimum value is when $x=\dfrac{1}{x}$ i.e. at $x=1$ If…
1
vote
0 answers

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 $$u^av^b\leq au+bv \tag{$*$}$$ for positive numbers $a,b,u,v$ with $a+b=1$…
KHOOS
  • 407
1
vote
1 answer

Prove $\left(\frac{a+b+c}{3}\right)^p\leq \frac{a^p+b^p+c^p}{3}$.

Prove: Let $p$ be an integer greater than $1$. Suppose $a,b,c$ be positive real numbers. Then $\left(\frac{a+b+c}{3}\right)^p\leq \frac{a^p+b^p+c^p}{3}$. By AM-GM, I get $\frac{a+b+c}{3}\geq (abc)^{1/3}$ and $\frac{a^p+b^p+c^p}{3}\geq…
sunny
  • 133
1
vote
1 answer

Hard AM-GM Inequality

Find the sum of all positive integers $n,$ where the inequality $\sqrt{a + \sqrt{b + \sqrt{c}}} \ge \sqrt[n]{abc}$ holds for all nonnegative real numbers $a,$ $b,$ and $c.$ I tried squaring both sides but I'm not sure how to continue and apply the…
AW23
  • 77
1
vote
1 answer

Proving the inequality using am gm theorem

I was asked to prove that $a^3+b^3 \le (a^2+b^2)(a^4+b^4)$ I expanded the rhs and used am gm and got $a^6+a^2b^4+a^4b^2+b^6 \ge 4a^3b^3$ and struck.i think I lack intuition or I don't have enough experience.Any hints
1
vote
1 answer

Find $a^3 + b^3 +c^3, $ given $a+b+c=12$ and $a^3 \cdot b^4 \cdot c^5 = 0.1 \cdot (600)^3$

$a+b+c=12$ and $a^3 \cdot b^4 \cdot c^5 = 0.1 \cdot (600)^3$. Find $a^3 + b^3 +c^3 = ?$ My approach is to use AM-GM inequality. Is it correct?
Mark 7
  • 43
1
vote
1 answer

why is Arithmetic mean minimum when all terms are equal

I am at a beginner's level (graduation 1st Year) going through the topic of university level inequalities for the first time. Read this recently:- "If $x_1,x_2,x_3 ,\dots,x_n$ are $n$ positive real numbers such that $x_1+x_2+\dots+x_n$ is a…
1
2 3