# 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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### Find the maximum value of $ab+bc+cd+de+ef+fa,$ given that $a+b+c+d+e+f=1$

After looking at this post, I framed the following question. If $a,b,c,d,e,f$ are positive real numbers such that $a+b+c+d+e+f=1$, then find the minimum and maximum value of $ab+bc+cd+de+ef+fa$ My ...
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### Prove or disprove that the inequality is valid if $x,y,z,u$ are positive numbers and $x+y+z+u=2$.

Prove or disprove that the inequality $$\dfrac{x^2}{\left(x^2+1\right)^2}+\dfrac{y^2}{\left(y^2+1\right)^2}+\dfrac{z^2}{\left(z^2+1\right)^2}+\dfrac{u^2}{\left(u^2+1\right)^2} \leq \dfrac{16}{25}$$ ...
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### To prove $1^1\cdot2^2\cdot 3^3...\cdot n^n<(\frac{2n+1}{3})^{\frac{n(n+1)}{2}}$

So we have to prove the following for $n\in N$ $$1^1\cdot 2^2\cdot 3^3...\cdot n^n<\left(\frac{2n+1}{3}\right)^{\frac{n(n+1)}{2}}$$ So I used concept of weighted means (arithmetic and geometric) ...
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### 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 ...
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### Prove that the inequality is valid if $x,y,z$ are positive numbers and $xyz=1.$

Is given that $x,y,z$ are positive numbers and $xyz=1$, prove that $$\dfrac{\dfrac{1}{x}}{\sqrt{z^2+1}}+\dfrac{\dfrac{1}{y}}{\sqrt{x^2+1}}+\dfrac{\dfrac{1}{z}} {\sqrt{y^2+1}}>\sqrt{2}.$$ What have ...
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### Help w/$(\frac{a}{b})^4+(\frac{b}{c})^4+(\frac{c}{d})^4+(\frac{d}{e})^4+(\frac{e}{a})^4\ge\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{e}{d}+\frac{a}{e}$ [closed]

How exactly do I solve this problem? (Source: 1984 British Math Olympiad #3 part II) \begin{equation*} \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}...
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### Find the minimum of $xy+yz+zx+\frac1x+\frac2y+\frac5z$ for $x, y, z > 0$

Let $x,y,z > 0$. What is the smallest value of the below expression? \begin{align*} xy+yz+zx+\frac1x+\frac2y+\frac5z \end{align*} I tried with Lagrange multiplier but I have not any constraint. 71 views

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### Prove that: $\sum_{1\le j\le n}{\frac{x_j}{\sum_{1\le i\le n} x_i-x_j}}\ge\frac{n}{n-1}$

$\color{red}{\textbf{Problem:}}$ Let, $x_i>0,1\le i\le n$, then Prove that: $$\sum_{1\le j\le n}{\frac{x_j}{\sum_{1\le i\le n} x_i-x_j}}\ge\frac{n}{n-1}$$ $\color{red}{\textbf{Proof:}}$ Using AM-...
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### Finding the minimum value of the expression $xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2$

From R. D. Sharma's Objective Mathematics, Given that $x + y + z = 1$, ($x,y,z$ are positive real numbers) find the minimum value of $$A = xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2$$ My attempt: By A.M.-G.M....
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### Geometric-arithmetic mean inequality applied to eigenvalues

I've applied the arithmetic-geometric mean inequality to the eigenvalues of a positive definite matrix $X$, so $det(X)^{1/n}≤tr(x)/n$. Now I would like to show when equality holds. I already found out ...
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### Prove $\frac{(c-a)^{2}}{6c} \le \frac{a+b+c}{3} - \frac{3}{ 1/a + 1/b+ 1/c}$

Given real numbers $c \ge b \ge a>0$, prove that $$\frac{(c-a)^{2}}{6c} \le \frac{a+b+c}{3} - \frac{3}{ \frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$$ *using well-known inequality Other solution ...
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Suppose $p_m \geq 0$ and $\sum_{m \in \mathbf{Z}} p_m =1 .$ That is $p$ is a probability measure on integers. Then how can I show (is it true) that  \sum_{m \in \mathbf{Z}} (p_m + p_{m+1} + p_{m+2})^...
Doing some calculus papers before going university and I found this question: Find $\lim_{x\to\infty} \left[\frac{1}{3} \left(3^\frac{1}{x} + 8^\frac{1}{x} + 9^\frac{1}{x} \right)\right]^x$ My ...