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
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the maximum value of $xy/(x+2y)$, given $x^2+4y^2+2xy=1, x>0, y>0$

My thought is $\frac{xy}{x+2y} = \frac{1}{2}\frac{x*2y}{x+2y}\le \frac{1}{8}\frac{(x+2y)^2}{x+2y}=\frac{x+2y}{8}$. I don't know how to proceed. Please use AM-GM Inequality only.
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Show that $\sqrt{\frac{a}{b+c}+ \frac{b}{c+a}}+ \sqrt{\frac{b}{c+a}+ \frac{c}{a+b}} + \sqrt{\frac{c}{a+b}+ \frac{a}{b+c}} \ge 3.$

Suppose that $a,b,c>0$, show that $$\sqrt{\frac{a}{b+c}+ \frac{b}{c+a}}+ \sqrt{\frac{b}{c+a}+ \frac{c}{a+b}} + \sqrt{\frac{c}{a+b}+ \frac{a}{b+c}} \ge 3.$$ Appreciate any advice!
Steven Lu
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AM-GM inequality involving $a^n + b^n$

I'm currently solving problems involving AM-GM and I am stuck on this problem. Prove that if $a,b,c>0$ and $n,k \in \mathbb{Z}^{+}, n>k$ that $a^n + b^n \geq a^{n-k}b^{k} + a^{k}b^{n-k}$ I've attempted the case when $n=3$ and found that $a^3 + b^3…
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If $a$, $b$, and $c$ are positive numbers, then $\sqrt{a^2+ab+b^2}+\sqrt{a^2+ac+c^2}+\sqrt{b^2+bc +c^2}\ge\sqrt{3}(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}).$

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 equality occur?
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A friends proof of AM-GM inequality

We are supposed to prove that $ \frac{\sum_{k=1}^{n}x_k}{n}\geq (\prod_{k=1}^{n} x_k)^{1/n}$, when $x_i\geq 0$ for all $n\in\mathbb{N}$. My friend proved it for $n=2$, $n=3$ and then for an arbitrary integer $n$. Now I only think the proof for an…
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Revisit a familiar 3-variable inequality .

If $x > 0, y > 0, z > 0$ and $x+y+z = 1$ then prove $$\dfrac{x}{2x+1}+\dfrac{y}{3y+1}+\dfrac{z}{6z+1}\le \dfrac{1}{2}$$ I saw this post half an hour ago and after coming up with a proof, the post was deleted. So I thought maybe OP of that post…
user899577
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If $s=a+b+c,p=ab+bc+ac, r=abc, s^2-2p+r=4$ and $a,b,c>0,$ prove that $2sp+15r^2\geq 33r$

If $s=a+b+c$,$p=ab+bc+ac$, $r=abc$, $s^{2}-2p+r=4$ and $a,b,c>0,$ prove that $$2sp+15r^{2}\geq 33r.$$ I tried to use some inequalities from here, but nothing works. Suggestions ?
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Solve the inequality $a^3/(2b^3+ab^2)+b^3/(2c^3+bc^2)+c^3/(2d^3+cd^2)+d^3/(2a^3+da^2)\geq 4/3$ for $a,b,c>0$.

Solve the inequality $\frac{a^3}{2b^3+ab^2}+\frac{b^3}{2c^3+bc^2}+\frac{c^3}{2d^3+cd^2}+\frac{d^3}{2a^3+da^2}\geq \frac{4}{3}$ for $a,b,c>0$. I tried to use the inequality $a^3+2b^3\geq 3ab^2$, etc., adding $1$ to each fraction. I tried with…
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Prove the following inequality using the AM-GM Mean

Let $x_0>x_1>x_2>\ldots>x_n$ be real numbers. Prove $\displaystyle x_{0}+\frac{1}{x_{0}-x_{1}}+\frac{1}{x_{1}-x_{2}}+\ldots+\frac{1}{x_{n-1}-x_{n}}\geq x_{n}+2n$. I thought I would try and deal with the denominators by letting…
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An inequality which I got confused on.

Prove that - $$3(x^2-x+1)(y^2-y+1) \geqq 2(x^2y^2 -xy+1)$$ My Approach I tried to solve this inequality using the A.M. - G.M. but I wasn't able to get the solution using it. I would appreciate if someone shows me the solution using the A.M. - G.M.…
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Inequality Problem using the A.M. - G.M. inequality

I request everyone to please post the solution to this problem using the A.M. - G.M. Inequality. The problem is as follows - Problem - Let $a,b,c$ be positive real numbers smaller than $1$ such that $abc=0.5$ Prove…
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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 $$x^\lambda y^{1-\lambda} \leq \lambda x+ (1-\lambda)y$$ But I know the…
user444830
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complex AM GM inequality, probably based on weighted means

I came across a question where i had to find the maximum value of a²b³c², and was just provided with the value of a+b+c which was 3 While trying to solve this i felt that i could use weighted means, which gave the maximum value of a²b³c² as…
makra
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Possible flaw in AM GM inequality

AM GM inequality states that for real positive numbers x,y $$x + y \geq 2(xy)^{1/2}$$ You would get the least value for x = y We can also write this as, $$\frac x3 + \frac x3 + \frac x3 + y \geq…