Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [uvw]

The uvw method is a very useful method for the proof of polynomial inequalities with three variables. Sometimes it works for more variables as well. This tag should be used for questions that could be tackled with this method, or questions about the method itself.

Let say we need to prove that $P\geq0$, where $P$ is a symmetrical polynomial over $\mathbb R$ of variables $a$, $b$ and $c$.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.

Hence, there is polynomial $f$ with three variables for which $f(u,v^2,w^3)=P(a,b,c)$.

We have the following statements.

  1. If $f$ is an increasing function of $w^3$ then for the proof of $P\geq0$ it's enough to prove it for an equality case of two variables.
  1. If $f$ is a decreasing function of $w^3$ then for the proof of $P\geq0$ it's enough to prove it for an equality case of two variables.
  1. If $f$ is a concave function of $w^3$ then for the proof of $P\geq0$ it's enough to prove it for an equality case of two variables.

For non-negative variables we have the following statements.

  1. If $f$ is an increasing function of $w^3$ then for the proof of $P\geq0$ it's enough to prove it for an equality case of two variables and for $w^3=0$.
  1. If $f$ is a decreasing function of $w^3$ then for the proof of $P\geq0$ it's enough to prove it for an equality case of two variables.
  1. If $f$ is a concave function of $w^3$ then for the proof of $P\geq0$ it's enough to prove it for an equality case of two variables.
  1. If $f$ is an increasing function of $v^2$ then for the proof of $P\geq0$ it's enough to prove it for an equality case of two variables.
  1. If $f$ is a decreasing function of $v^2$ then for the proof of $P\geq0$ it's enough to prove it for an equality case of two variables.
  1. If $f$ is a concave function of $v^2$ then for the proof of $P\geq0$ it's enough to prove it for an equality case of two variables.
  1. If $f$ is an increasing function of $u$ then for the proof of $P\geq0$ it's enough to prove it for an equality case of two variables.
  1. If $f$ is a decreasing function of $u$ then for the proof of $P\geq0$ it's enough to prove it for an equality case of two variables.
  1. If $f$ is a concave function of $u$ then for the proof of $P\geq0$ it's enough to prove it for an equality case of two variables.
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Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\frac{3}{\sqrt{7}}$

Let $a,b,c>0$ such that $$\dfrac{1}{a^2+2}+\dfrac{1}{b^2+2}+\dfrac{1}{c^2+2}=\dfrac{1}{3}.$$ Show that $$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\dfrac{3}{\sqrt{7}}.$$ My try:…
math110
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How prove this inequality $\sum\limits_{cyc}\frac{x+y}{\sqrt{x^2+xy+y^2+yz}}\ge 2+\sqrt{\frac{xy+yz+xz}{x^2+y^2+z^2}}$

let $x,y,z$ are postive numbers,show that $$\dfrac{x+y}{\sqrt{x^2+xy+y^2+yz}}+\dfrac{y+z}{\sqrt{y^2+yz+z^2+zx}}+\dfrac{z+x}{\sqrt{z^2+zx+x^2+xy}}\ge 2+\sqrt{\dfrac{xy+yz+xz}{x^2+y^2+z^2}}$$ My try: Without loss of generality，we assume that…
math110
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How prove this inequality $\frac{2}{(a+b)(4-ab)}+\frac{2}{(b+c)(4-bc)}+\frac{2}{(a+c)(4-ac)}\ge 1$

let $a,b,c>0$,and such $a+b+c=3$, show that $$\dfrac{2}{(a+b)(4-ab)}+\dfrac{2}{(b+c)(4-bc)}+\dfrac{2}{(a+c)(4-ac)}\ge 1$$ I think this inequality use this $$ab\le\dfrac{(a+b)^2}{4}$$
math110
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How to prove this inequality? $ab+ac+ad+bc+bd+cd\le a+b+c+d+2abcd$

let $a,b,c,d\ge 0$,and $a^2+b^2+c^2+d^2=3$,prove that $ab+ac+ad+bc+bd+cd\le a+b+c+d+2abcd$ I find this inequality are same as Crux 3059 Problem.
math110
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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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How prove this $x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$

Question: let $x,y,z>0$ and such $xyz=1$, show that $$x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$$ My idea: use AM-GM inequality $$x^3+x^3+1\ge 3x^2$$ $$y^3+y^3+1\ge 3y^2$$ $$z^3+z^3+1\ge 3z^2$$ so $$2(x^3+y^3+z^3)+3\ge 3(x^2+y^2+z^2)$$ But this is not my…
math110
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Prove $\frac{1}{2a+2bc+1} + \frac{1}{2b+2ca+1} + \frac{1}{2c+2ab+1} \ge 1$

If $a,b$ and $c \ge 0$ and $ab + bc + ca = 1$, prove that the following inequality holds: $$\frac{1}{2a+2bc+1} + \frac{1}{2b+2ca+1} + \frac{1}{2c+2ab+1} \ge 1$$ I've tried two aproaches, but it seems like both doesn't work. Here they…
Stefan4024
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Hard inequality for positive numbers

The problem is to prove that for $a,b,c>0$ we have $$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{9abc}{4(a^3+b^3+c^3)}\geq \frac{15}{4}.$$ I have tried to use Bergstrom/Engel inequality to write, for example,…
JohnnyC
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On the Abstract Concreteness Method (bka $ABC-$Method).

I was reading Zdravko Cvetkovski's excellent book Inequalities: Theorems, Techniques, and selected problems, when I arrived at the $16$th chapter: the $ABC-$Method. I had some questions related to this topic, which I could post separately; yet,…
Dr. Mathva
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Prove that $({a\over a+b})^3+({b\over b+c})^3+ ({c\over c+a})^3\geq {3\over 8}$

Let $a,b,c$ be positive real numbers. Prove that $$\Big({a\over a+b}\Big)^3+\Big({b\over b+c}\Big)^3+ \Big({c\over c+a}\Big)^3\geq {3\over 8}$$ If we put $x=b/a$, $y= c/b$ and $z=a/c$ we get $xyz=1$ and $$\Big({1\over 1+x}\Big)^3+\Big({1\over…
nonuser
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prove this inequality with $63$

Let $x,y,z,w>0$, and such $x^2+y^2+z^2+w^2=1$. show that $$x+y+z+w+\dfrac{1}{63xyzw}\ge\dfrac{142}{63}\tag{1}$$ I know $$x^2+y^2+z^2+w^2\ge 4\sqrt[4]{x^2y^2z^2w^2}\Longrightarrow xyzw\le \dfrac{1}{16}$$ so we…
math110
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Prove that $\sum\limits_{cyc}\frac{a}{a+b}\geq1+\frac{3\sqrt[3]{a^2b^2c^2}}{2(ab+ac+bc)}$

Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that: $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq1+\frac{3\sqrt[3]{a^2b^2c^2}}{2(ab+ac+bc)}$$ I tried C-S, uvw, BW and more, but without success.
Michael Rozenberg
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How prove that $xyz+\sqrt{x^2y^2+y^2z^2+x^2z^2}\ge \frac{4}{3}\sqrt{xyz(x+y+z)}$

let $x,y,z>0$,and such that $x^2+y^2+z^2=1$,prove that $$xyz+\sqrt{x^2y^2+y^2z^2+x^2z^2}\ge \dfrac{4}{3}\sqrt{xyz(x+y+z)}$$ Does this have a nice solution? Thank you everyone.
math110
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It may be a strengthening form of mean inequality

Let $a_{i}>0,i=1,2,\cdots,n>2$,show that following inequality $$\prod_{i=1}^{n}\left(\dfrac{\displaystyle\sum_{j\neq i}a_{j}}{n-1}\right)^{n-1}\ge\left(\dfrac{\displaystyle\sum_{i=1}^{n}\prod_{j\neq i}a_{j}}{n}\right)^n$$ I have found that this…
math110
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