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Questions tagged [uvw]

This is a very useful method for the proof of polynomial inequalities with three variables. Sometimes it works for more variables.

4
votes
3answers
6k views

How to prove this inequality $\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}}$

How to prove this inequality $$\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}} ?$$ Thanks
20
votes
4answers
902 views

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 $$ab\le\dfrac{(a+b)^2}{4}$$
8
votes
3answers
666 views

Hard inequality $ (xy+yz+zx)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\ge\frac{9}{4} $

I need to prove or disprove the following inequality: $$ (xy+yz+zx)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\ge\frac{9}{4} $$ For $x,y,z \in \mathbb R^+$. I found no counter ...
7
votes
4answers
380 views

Prove $\sum{\frac{1}{(x+2y)^2}} \geq\frac{1}{xy+yz+zx}$

Let $x,y,z>0$: Prove that: $\frac{1}{(x+2y)^2}+\frac{1}{(y+2z)^2}+\frac{1}{(z+2x)^2} \geq\frac{1}{xy+yz+zx}$ I tried to apply Cauchy - Schwarz's inequality but I couldn't prove this inequality!
6
votes
3answers
322 views

How prove that $10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$

let $a,b,c\ge 0$, such that $a+b+c=1$, prove that $$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$$ This problem is simple as 2005, china west competition problem $$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\ge ...
5
votes
4answers
199 views

Find the maximum value of the expression ${\frac {x}{1+x^2}} + {\frac {y}{1+y^2}}+{\frac {z}{1+z^2}}$. [duplicate]

Find the maximum value of the expression : $${\frac {x}{1+x^2}} + {\frac {y}{1+y^2}}+{\frac {z}{1+z^2}}$$ where $x,y,z$ are real numbers satisfying the condition that $x+y+z=1$. Taking $x=y=z=\...
6
votes
3answers
787 views

Proving inequality $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4$

I started to study inequalities - I try to solve a lot of inequlites and read interesting .solutions . I have a good pdf, you can view from here . The inequality which I tried to solve and I didn't ...
2
votes
2answers
147 views

Prove $(1+a+b)(1+b+c)(1+c+a)\ge 9(ab+bc+ca)$

How one can prove the following. Let $a$, $b$ and $c$ be non-negative real numbers. Then the inequality holds: $(1+a+b)(1+b+c)(1+c+a)\ge9(ab+bc+ca).$ WLOG one can assume that $0\le a\le b\le c$. ...
13
votes
2answers
1k views

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....
8
votes
2answers
228 views

show that inequality $(x-3)^4+(y-3)^4+(z-3)^4\ge 193$

Let $x,y,z\in R$,and such $$xy+yz+xz=-1$$ show that $$(x-3)^4+(y-3)^4+(z-3)^4\ge 193$$ it seem use Cauchy-Schwarz inequality to solve it?But I try sometime can't get this answer,even now I can't find ...
9
votes
2answers
419 views

If the sum of cubes of $a,b,c,d$ is $1$, then $\frac{1}{1-bcd}+\frac{1}{1-cda}+\frac{1}{1-dab}+\frac{1}{1-abc}\le \frac{16}{3}$

$a,b,c,d>0$ satisfying $a^3+b^3+c^3+d^3=1$. Prove $$\frac{1}{1-bcd}+\frac{1}{1-cda}+\frac{1}{1-dab}+\frac{1}{1-abc}\le \frac{16}{3}$$ I tried to go the normal way, by Cauchy-Schwarz, but that ...
5
votes
2answers
257 views

Prove $\frac{x+y+z}{3}+\frac{3}{\frac1x+\frac1y+\frac1z}\geq5\sqrt[3]{\frac{xyz}{16}}$

Let $x,y,z>0$. Prove that $$\frac{x+y+z}{3}+\frac{3}{\frac1x+\frac1y+\frac1z}\geq5\sqrt[3]{\frac{xyz}{16}}$$ On the left-hand side we have arithmetic and harmonic means, while on the right ...
15
votes
4answers
689 views

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(...
9
votes
2answers
179 views

I conjecture this inequality $\sqrt[4]{\frac{(xy+yz+xz)(x^2+y^2+z^2)}{9}}\ge\sqrt[3]{\frac{(x+y)(y+z)(z+x)}{8}}$

Let $x,y,z>0$,prove or disprove $$\sqrt[4]{\dfrac{(xy+yz+xz)(x^2+y^2+z^2)}{9}}\ge\sqrt[3]{\dfrac{(x+y)(y+z)(z+x)}{8}}$$ I tried many times ,use $$x^2+y^2+z^2\ge\dfrac{1}{3}(x+y+z)^2$$ and kown $$9(...
3
votes
3answers
222 views

Finding the maximum of $p^3 + q^3 +r^3 + 4pqr$

$p$,$q$,$r$ are $3$ non-negative real numbers less than or equal to $1.5$ such that $p+q+r = 3$, what will be the maximum of $p^3 + q^3 + r^3 + 4pqr$ ? I tried AM-GM on $p,q,r$ to get the maximum of $...
3
votes
1answer
239 views

If $a+b+c=3$ show $a^2+b^2+c^2 \leq (27-15\sqrt{3})\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$

I have become interested in constrained relations among simple cyclic sums involving three positive variables. By simple, I mean so simple that they are also fully symmetric. The "building blocks" of ...
1
vote
2answers
162 views

Find the minimum value of $P=\sum _{cyc}\frac{\left(x+1\right)^2\left(y+1\right)^2}{z^2+1}$

For $x>0$, $y>0$, $z>0$ and $x+y+z=3$ find the minimize value of $$P=\frac{\left(x+1\right)^2\left(y+1\right)^2}{z^2+1}+\frac{\left(y+1\right)^2\left(z+1\right)^2}{x^2+1}+\frac{\left(z+1\...
1
vote
5answers
111 views

If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$ [closed]

If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$ This can be easily done by calculas but is there any way to do do this by algebra
7
votes
1answer
224 views

$ \frac {3 x y z}{(x+y) (y+z) (z+x)} + \sum\limits_{cycl}^{} \left(\frac {x+y}{x+y+ 2 z}\right)^2 \ge \frac {9}{8}.$

For $x,y,z>0,$ I have to prove that $$ \frac {3 x y z}{(x+y) (y+z) (z+x)} + \sum\limits_{cycl}^{} \left(\frac {x+y}{x+y+ 2 z}\right)^2 \ge \frac {9}{8}.$$ I tried to use $$ \sqrt {\frac {1}{3} \...
2
votes
1answer
85 views

An inequality with condition

I have a new inequality this is the following : Let $x,y,z$ be real strictly positive number such as : $$-2 = - x y z + x + y + z $$ Then we have : $$\sqrt{\frac{1}{x}}+\sqrt{\frac{1}{y}}+\...