# 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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problem: a,b,c>0 and $abc=1$ prove that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq a+b+c$ my attempt: $LHS=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=(a+b+c) \frac{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}{a+b+... 5 votes 1 answer 738 views ### Weighted AM-GM Inequality I was tried to learn some tools on inequalities, and I learn the "Weighted AM-GM Inequality" in a small book: in this book there is an exercise and a solution, I did my attempt and it was ... 0 votes 2 answers 82 views ### prove that :$(1+\frac{1}{n})^k<1+\frac{k}{n}+\frac{k^2}{2n^2}$. if$ n,k \in \mathbb{N^*}$and$(k-1)^2<n$prove that :$(1+\frac{1}{n})^k<1+\frac{k}{n}+\frac{k^2}{2n^2}$. my attempt:$(1+\frac{1}{n})^k+1=(1+\frac{1}{n}).(1+\frac{1}{n})^{k-1}+1\leq \sqrt{(1+\... 94 views

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### Finding a limit using AM-GM?

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 ...
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### Find number of continuous functions satisfying the equation $4\int_{0}^{\frac{3}{2}}f(x)dx+125\int_{0}^{\frac{3}{2}}\frac{dx}{\sqrt{f(x)+x^2}}=108$

The number of continuous functions $f:\left[0,\frac{3}{2}\right]\rightarrow (0,\infty)$ satisfying the equation$$4\int_{0}^{\frac{3}{2}}f(x)dx+125\int_{0}^{\frac{3}{2}}\frac{dx}{\sqrt{f(x)+x^2}}=108$$ ...
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### Let $x,y,z\in[0,1]$. Find the maximum value of $\sqrt{|x-y|}+\sqrt{|y-z|}+\sqrt{|z-x|}$. [duplicate]

Let $x,y,z\in[0,1]$. Then find the maximum value of $\sqrt{|x-y|}+\sqrt{|y-z|}+\sqrt{|z-x|}$. Now the given answer is $2\sqrt{2}$ but I am not able to obtain the corresponding values of $x,y,z$. ...
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### How can i find minimum value of this inequality

$a_1,a_2,...,a_n$ are 8 distinct positive integers. $b_1,b_2,...,b_n$ are another 8 distinct positive integers ($a_i,b_j$ are not necessarily y distinct for $i, j = 1, 2, ...8$).Enter the smallest ...
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