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 $w=\sum_{i=1}^nw_i$.

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Proofs of AM-GM inequality

The arithmetic - geometric mean inequality states that $$\frac{x_1+ \ldots + x_n}{n} \geq \sqrt[n]{x_1 \cdots x_n}$$ I'm looking for some original proofs of this inequality. I can find the usual proofs on the internet but I was wondering if someone…
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Is this continuous analogue to the AM–GM inequality true?

First let us remind ourselves of the statement of the AM–GM inequality: Theorem: (AM–GM Inequality) For any sequence $(x_n)$ of $N\geqslant 1$ non-negative real numbers, we have $$\frac1N\sum_k x_k \geqslant \left(\prod_k x_k\right)^{\frac1N}$$ It…
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New bound for Am-Gm of 2 variables

Today I'm interested by the following problem : Let $x,y>0$ then we have : $$x+y-\sqrt{xy}\leq\exp\Big(\frac{x\ln(x)+y\ln(y)}{x+y}\Big)$$ The equality case comes when $x=y$ My proof uses derivative because for $x\geq y $ the function…
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Can we prove AM-GM Inequality using these integrals?

I came across these two results recently: $$ \int_a^b \sqrt{\left(1-\dfrac{a}{x}\right)\left(\dfrac{b}{x}-1\right)} \: dx = \pi\left(\dfrac{a+b}{2} - \sqrt{ab}\right)$$ $$ \int_a^c \sqrt[3]{\left|…
Amioun
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Inequality with five variables

Let $a$, $b$, $c$, $d$ and $e$ be positive numbers. Prove that: $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+e}+\frac{e}{e+a}\geq\frac{a+b+c+d+e}{a+b+c+d+e-3\sqrt[5]{abcde}}$$ Easy to show that…
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Is the AM-GM inequality the only obstruction for getting a specific sum and product?

This might be silly, but here it goes. Let $P,S>0$ be positive real numbers that satisfy $\frac{S}{n} \ge \sqrt[n]{P}$. Does there exist a sequence of positive real numbers $a_1,\dots,a_n$ such that $S=\sum a_i,P=\prod a_i$? Clearly, $\frac{S}{n}…
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Prove $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$

Let $a,b,c$ are non-negative numbers, such that $a+b+c = 3$. Prove that $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$ Here's my idea: $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$ $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(ab + bc +…
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Prove $\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$

If $a,b,c$ are non-negative numbers and $a+b+c=3$, prove that: $$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6.$$ Here's what I've tried: Using Cauchy-Schawrz I proved that: $$(3a + b^3)(3 + 1) \ge (3\sqrt{a} +…
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Prove the inequality $\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$ with the constraint $abc=1$

If $a,b,c$ are positive reals such that $abc=1$, then prove that $$\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$$ I tried substituting $x/y,y/z,z/x$, but it didn't help(I got the reverse inequality). Need some stronger…
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Finding maxima of a function $f(x) = \sqrt{x} - 2x^2$ without calculus

My question is how to prove that $f(x) = \sqrt x - 2x^2$ has its maximum at point $x_0 = \frac{1}{4}$ It is easy to do that by finding its derivative and setting it to be zero (this is how I got $x_0 = \frac{1}{4}$). But the task is to do that…
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Combined AM GM QM inequality

I came across this interesting inequality, and was looking for interesting proofs. $x,y,z \geq 0$ $$ 2\sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}}+3\sqrt [3]{xyz}\leq 5\left(\frac{x+y+z}{3}\right) $$ Addendum. In general, when is $$…
picakhu
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The inequality $\,2+\sqrt{\frac p2}\leq\sum\limits_\text{cyc}\sqrt{\frac{a^2+pbc}{b^2+c^2}}\,$ where $0\leq p\leq 2$ is: Probably true! Provably true?

Let $p$ be a positive parameter in the range from $0$ to $2$. Can one prove that $$2 +\sqrt{\frac p2} \;\leqslant\;\sqrt{\frac{a^2 + pbc}{b^2+c^2}} \,+\,\sqrt{\frac{b^2 +pca}{c^2+a^2}}\,+\,\sqrt{\frac{c^2 +pab}{a^2+b^2}}\quad?\tag{1}$$ Where…
Hanno
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Prove $(a_1+b_1)^{1/n}\cdots(a_n+b_n)^{1/n}\ge \left(a_1\cdots a_n\right)^{1/n}+\left(b_1\cdots b_n\right)^{1/n}$

consider positive numbers $a_1,a_2,a_3,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$. does the following in-equality holds and if it does then how to prove it $\left[(a_1+b_1)(a_2+b_2)\cdots(a_n+b_n)\right]^{1/n}\ge \left(a_1a_2\cdots…
Mia
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How to prove the inequality $ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$

How to prove the inequality $$ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$$ for $a,b,c>0$ and $abc=1$? I have tried prove $\frac{a}{\sqrt{1+a}}\ge \frac{3a+1}{4\sqrt{2}}$ Indeed,$\frac{{{a}^{2}}}{1+a}\ge…
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