# 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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### Hint to prove an inequality

some help to prove this inequality : $$\forall \,\, a,b,c \geq 0 \\ \frac{a}{a^2+b^2 +2} +\frac{b}{b^2+c^2 +2} + \frac{c}{c^2+a^2 +2} \leq \frac{3}{4}$$ Thank
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### Prove or disprove that the inequality is true if $xyz=1$ and $xyz=8^3$.

Prove that the inequality $$\dfrac{1}{\sqrt{x+1}}+\dfrac{1}{\sqrt{y+1}}+\dfrac{1}{\sqrt{z+1}} \geq 1$$ is true if $x, y, z>0$ and $1)$ $xyz=1$ $2)$ $xyz=8^3.$ For the first case it turns out to ...
1 vote
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### Prove $2(a^2+b^2+c^2)+\frac{4}{3}\sum\limits_{\mathrm{cyc}}\frac{1}{a^2+1} \geq 5$ for $ab+bc+ac=1$

The positive real numbers a, b, and c satisfy ab+bc+ac=1. Prove that the inequality $$2 \left(a^2+b^2+c^2\right)+\dfrac{4}{3}\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right) \geq 5$$ ...
1 vote
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### Prove $\sqrt{\frac{x}{\left(x+y\right)\left(1+y\right)}}+\sqrt{\frac{y}{\left(x+y\right)\left(1+x\right)}}+\sqrt{\frac{xy}{(1+x)(1+y)}}>1$

For $x, y, z > 0$, prove that $$\sqrt{\dfrac{x}{\left(x+y\right)\left(1+y\right)}}+\sqrt{\dfrac{y}{\left(x+y\right)\left(1+x\right)}}+\sqrt{\dfrac{xy}{\left(1+x\right)\left(1+y\right)}}>1.$$ I ...
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### Given the sum of $k$ numbers, what is the maximum value of their product? [duplicate]

Let $x_1,x_2,\ldots,x_k\in\mathbb{N}.$ Given that $x_1+x_2+\ldots+x_k=100,$ what is the maximum value of $P=x_1x_2\ldots x_k,$ and for what value of $k$ does this maximum occur? This is a question ...
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### If $x,y,z\in\mathbb R^+$ are in Harmonic Progression, prove that $z\cdot e^{x-y}+x\cdot e^{z-y}\ge\frac{2xz}y$

If $x,y,z\in\mathbb R^+$ are in Harmonic Progression, prove that $z\cdot e^{x-y}+x\cdot e^{z-y}\ge\frac{2xz}y$ Given, $\frac1y-\frac1x=\frac1z-\frac1y\implies z(x-y)=x(y-z)$ I tried AM-GM inequality ...
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### AM-GM Inequality. Let a, b, c be positive real numbers. Prove that $\frac {a+b+c}{3} \cdot (a^2+b^2+c^2) \ge a^2b + b^2c + c^2a$.

Prove that $$\frac {a+b+c}{3} \cdot \frac{a^2+b^2+c^2}{3} \ge \frac{a^2b + b^2c + c^2a}{3}$$. given that a,b,c are positive real numbers. Solve only using AM-GM. So far I have tried expanding LHS, ...
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### bounding $(|\alpha|+1)^2+(x+|\alpha|)^2$

Let $x>0$ and $\alpha\in\mathbb{R}$ be a 'parameter'. If possible, I would like to find an upper bound for the quantity $(|\alpha|+1)^2+(x+|\alpha|)^2$. All ideas are welcome. If there's an ...
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### Three-Variable Inequality

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 ...
1 vote
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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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Prove or disprove that the inequality $$\dfrac{1}{\sqrt{1+x}}+\dfrac{1}{\sqrt{1+y}}+\dfrac{1}{\sqrt{1+z}} \geq 1$$ is valid if $x,y,z$ are positive numbers and $$xyz=1.$$ My solution is: Let $$x=\... 1 vote 1 answer 91 views ### Show that x^{\frac1x}<1.5 for x \in \mathbb R Well, I've proven that :$$x^{\frac1x}<1.5$$for x \in \mathbb R^+, or more specifically I've shown that the maximum value of the expression is 1.44... for x = e. But I used calculus (finding ... 2 votes 1 answer 142 views ### Proof of a tighter inequality than Cauchy-Schwarz inequality A few days ago, I found this post which discuss about some tighter versions of Cauchy-Schwarz (or some might prefer the name AM-GM) inequality. User @Michael Rozenberg proposed a very interesting ... 7 votes 3 answers 163 views ### To prove 1^1\cdot2^2\cdot 3^3...\cdot n^n<(\frac{2n+1}{3})^{\frac{n(n+1)}{2}}  So we have to prove the following for n\in N$$1^1\cdot 2^2\cdot 3^3...\cdot n^n<\left(\frac{2n+1}{3}\right)^{\frac{n(n+1)}{2}} $$So I used concept of weighted means (arithmetic and geometric) ... 0 votes 1 answer 38 views ### Does the AM-GM inequality not hold if there is a variable on the GM side? To find the minimum value of x+\dfrac{1}{x}, (x>0) we can use the AM-GM inequality to say that$$\dfrac{x+\dfrac{1}{x}}{2}\geq \sqrt{x\cdot \dfrac{1}{x}}$$or$$x+\dfrac{1}{x}\geq 2$$The ... 1 vote 1 answer 83 views ### Prove that the inequality is valid if x,y,z are positive numbers and xyz=1. Is given that x,y,z are positive numbers and xyz=1, prove that$$\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}.$$What have ... 1 vote 2 answers 75 views ### Help w/(\frac{a}{b})^4+(\frac{b}{c})^4+(\frac{c}{d})^4+(\frac{d}{e})^4+(\frac{e}{a})^4\ge\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{e}{d}+\frac{a}{e} [closed] How exactly do I solve this problem? (Source: 1984 British Math Olympiad #3 part II) \begin{equation*} \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}... 1 vote 0 answers 26 views ### Sequence of geometric mean subtracted by arithmetic mean Let a_1,a_2,a_3,\dots be a sequence of positive numbers. Define$$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{$*$...
Let $a$ and $b$ be two positive numbers such that $a+b=1$. I am supposed to show that $u^av^b\leq au+bv$ for all positive $u$ and $v$. It is known that $\ln x \leq x$ for all positive $x$, so I ...