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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.

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2answers
66 views

Find the minimum and maximum values of $P=\left( 6-a^2-b^2-c^2\right)\left(2-abc\right)$

Let $a,b,c$ be non-negative real numbers such that $c \geq 1$ and that $a+b+c=2$. Find the minimum and maximum values of $$P=\left( 6-a^2-b^2-c^2\right)\left(2-abc\right)$$ To find the minimum of $P$ ...
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1answer
64 views

An inequality with the $constant= \frac{1}{2}+ \frac{5}{18}\,\sqrt{3}$

Given $a,\,b,\,c> 0$ such that$:$ $a+ b+ c= 3$$.$ Prove$:$ $$\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}\geqq \left ( \frac{1}{2}+ \frac{5}{18}\,\sqrt{3} \right )(\,a^{\,2}+ b^{\,2}+ c^{\,2}\,)$$ I find $...
1
vote
0answers
66 views

Find this inequality by finding the generalizating one too

Given $x,\,y,\,z$ such that $x,\,y,\,z\in [\,1,\,8\,]$$.$ Prove$:$ $$\frac{x}{y}+ \frac{y}{z}+ \frac{z}{x}\geqq \frac{2\,x}{y+ z}+ \frac{2\,y}{z+ x}+ \frac{2\,z}{x+ y}$$ By computer$,$ I have found ...
2
votes
3answers
53 views

Inequalities, but working around an absolute value

So what I want to prove is $$ |xy+xz+yz- 2(x+y+z) + 3| \leq |x^2+y^2+z^2-2(x+y+z)+3| $$ for $x,y,z\in \mathbb{R}$, and I'm aware that the RHS is just $|(x-1)^2+(y-1)^2+(z-1)^2|$. Now I'm able to ...
2
votes
2answers
66 views

Inequality with a+b+c=1 and $18(a^4+b^4+c^4)+6(a^2+b^2+c^2)+1\geq24(a^3+b^3+c^3)$

Let $a,b,c$ be reals with $a+b+c=1$. Show that : $$18(a^4+b^4+c^4)+6(a^2+b^2+c^2)+1\geq24(a^3+b^3+c^3).$$ I have tried to something like this: $$18a^4-24a^3+6a^2-12a+12\geq 0$$ $$18b^4-24b^3+6b^2-12b+...
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votes
3answers
286 views

$\sqrt{24ab+25}+\sqrt{24bc+25}+\sqrt{24ca+25}\geq 21$ if $a+b+c=ab+bc+ca$?

For $a,b,c>0 $ and $a+b+c=ab+bc+ca$ . Prove or disprove that : $\sqrt{24ab+25}+\sqrt{24bc+25}+\sqrt{24ca+25}\geq 21$ I checked in very many cases. Example :$c=1, a=2,b=\frac{1}{2}...$ then it’s ...
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2answers
61 views

Find min of $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+4\sqrt{2}\sqrt{\frac{ab+bc+ac}{a^2+b^2+c^2}}$ [closed]

Given the three real numbers a, b, c are not negative, in which at most some are equal to zero. Find min of $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+4\sqrt{2}\sqrt{\frac{ab+bc+ac}{a^2+b^2+c^2}}$ ...
3
votes
1answer
76 views

Prove $\Sigma_{cyc}(\frac{a}{b-c}-3)^4\ge193$

The inequality is expected original question of this MSE question. The exact statement is "If $a$, $b$ and $c$ are positive real numbers and none of them are equal pairwise, prove the following ...
2
votes
1answer
177 views

Prove $ \frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} + \frac{36}{a + b + c} \geq 20 $

Show that if $a,b,c > 0$, such that $ab + bc + ca = 1$, then the following inequality holds: $$ \frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} + \frac{36}{a + b + c} \geq 20 $$ What I ...
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3answers
265 views

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 $xyz=1$ ...
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1answer
70 views

How to prove the given inequality using CS or AM GM HM inequality [closed]

I was solving a problem on triangular inequalities and I have ended up with the following inequality required to be proven $$\sum_{cyc}\frac{x}{2x+y+z}\geq\frac{9\sqrt{3(x+y+z)xyz}}{4(x+y+z)^2},$$ ...
0
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2answers
87 views

UVW method application of basic theorem

Find the minimum and maximum value of $x+y+z+xy+yz+xz$ if $x^2+y^2+z^2 = 1$ I converted it to uvw, $3u+3v^2$ is the expression, $(3u)^2-2(3v^2)=1$, is the constraint. now I don't know what to do, ...
1
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1answer
172 views

If $x+y+z=3$ then $\sum x\sqrt{x^3+3y} \ge 6$

Let $x,y,z>0$ such that $x+y+z=3$. Prove that $$\sum x\sqrt{x^3+3y} \ge 6$$ This trying doesn't help. With Cauchy Schwarz $(\sum x\sqrt{x^3+3y})^2\geq \sum x^2\sum(x^3+3y) = (x^2 + y^2 + z^2)(x^...
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votes
2answers
207 views

Prove that: $xy\sqrt{z}+yz\sqrt{x}+zx\sqrt{y}\geq x+y+z$

Let $x$,$y$ and $z$ are positive and $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\leq 3$$ Prove that: $$xy\sqrt{z}+yz\sqrt{x}+zx\sqrt{y}\geq x+y+z$$ The things I have done so far $$3\geq \sum \limits_{cyc}\...
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2answers
193 views

Inequality with $abc=2$

Let $a,b,c$ be positive real numbers such that $abc=2$. Prove that $\frac{a+b+c}{4}\geq \sqrt[4]{\frac{a^2+b^2+c^2}{6}}$ Sorry, I don't have an idea for this problem.If I had, I would have showed them ...
1
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1answer
171 views

Prove that $\sqrt{1-x^2} + \sqrt{1-y^2} + \sqrt{1-z^2}\geq 4\sqrt{\frac{3-(x^2+y^2+z^2)}{5+x^2+y^2+z^2}}$.

If $x,y,z \in[0,1/2]$, with $x+y+z=1$, then prove that: $$\sqrt{1-x^2} + \sqrt{1-y^2} + \sqrt{1-z^2}\geq 4\sqrt{\frac{3-(x^2+y^2+z^2)}{5+x^2+y^2+z^2}}$$ OK so... I've tried to square the expression ...
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votes
1answer
121 views

How to prove this inequality ??? [closed]

Let $a,b,c$ are positive real numbers such that $a+b+c=3$. Prove that $$\sum\frac{(7a^{3}+3)(b+c)}{7a+3} \geq 6 $$ I try to prove $LHS \geq \sum\frac{9}{5}a+\frac{1}{5}$ but don't succeed
7
votes
1answer
223 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} \...
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2answers
109 views

What tools one should use for inequalities?

If $a,b,c>0$ prove that: $$\frac{1}{a+4b+4c}+\frac{1}{4a+b+4c}+\frac{1}{4a+4b+c}\leq \frac{1}{3\sqrt[3]{abc}}.$$ My first try was the following: $$\sum_{cyc}\frac{1}{a+4b+4c}\leq\sum_{cyc}\frac{1}{\...
0
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1answer
40 views

Double integral of $(y-2)^2$ over $\{(x,y)\in\mathbb{R}^2\colon x^2 +y^2 \le 6x\}$

Trying to solve this for hours... will appreciate any help $$\int\int_{D}^{} (y-2)^2dxdy$$ $$D=\{(x,y)| x^2+y^2\leq6x\}$$ Tried doing it in a few ways... doesn't seem to work out.
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1answer
61 views

An inequality with a condition

If $x,y,z>0$ and $\sum_{cyc}^{} x^2 (1-x) \ge 0 $, I have to prove that: $\sum_{cyc}^{} x^2 y^2 (1-xy) \ge 0 $ Then, if $\sum_{cyc}^{} x=3,$ then I have to prove $\sqrt [8] \frac {\sum_{cyc}^{} ...
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1answer
165 views

A tricky algebraic inequality

This is an old inequality but I haven't seen a satisfactory solution yet and am hoping someone here can provide one. There are a couple of brute force solutions but they provide no insight into the ...
11
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1answer
421 views

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{...
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2answers
610 views

Whenever $a+b+c=1$, $\frac{bc+a+1}{a^2+1} + \frac{ac+b+1}{b^2+1} + \frac{ab+c+1}{c^2+1} \le \frac{39}{10}$

Prove that for a + b + c =1 and a,b,c are positive real numbers, then $$\frac{bc+a+1}{a^2+1} + \frac{ac+b+1}{b^2+1} + \frac{ab+c+1}{c^2+1} \le \frac{39}{10}$$ My try: if one term is proven to be $\...
1
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1answer
68 views

show this $\prod(a^3+b^3+ab)^2\ge (\sum a^2b)^3( \sum ab^2)^3$

let $a,b,c>0$ show that :$$(a^3+b^3+abc)^2(b^3+c^3+abc)^2(c^3+a^3+abc)^2-(a^2b+b^2c+c^2a)^3\cdot(ab^2+bc^2+ca^2)^3\ge 0$$Maybe there's some kind of identity in there.so I use wolfampha Calculated ...
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1answer
60 views

Stronger than Nessbit

Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that: $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\frac{3\sqrt[3]{a^2b^2c^2}}{2(a^2+b^2+c^2)}\geq2\tag{1}$$ we know ...
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2answers
135 views

If $x+y+z=2$ prove that $\sum_{cyc}\frac{1}{\sqrt{x^2+y^2}}\ge2+\frac{1}{\sqrt{2}}$

Let $x,y,z$ be non-negative reals whose sum is $2$. Prove that $\frac{1}{\sqrt{x^2+y^2}}+\frac{1}{\sqrt{y^2+z^2}}+\frac{1}{\sqrt{z^2+x^2}}\ge2+\frac{1}{\sqrt{2}}$ I have tried bounding them up (...
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1answer
83 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}}+\...
1
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1answer
89 views

Prove this inequality $Σ_{cyc}\sqrt{\frac{a}{b+3c}}\ge \frac{3}{2}$ with $a;b;c>0$

Let $a,b,c>0$. Prove $$\sqrt{\frac{a}{b+3c}}+\sqrt{\frac{b}{c+3a}}+\sqrt{\frac{c}{a+3b}}\ge \frac{3}{2}$$ $A=\sqrt{\frac{a}{b+3c}}+\sqrt{\frac{b}{c+3a}}+\sqrt{\frac{c}{a+3b}}$ Holder: $A^2\cdot ...
3
votes
3answers
128 views

Prove $\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq 64abc$

Give $a, b, c$ be positive real numbers such that $a+ b+ c= 3$. Prove that: $$\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq 64abc$$ My try We have: $$\left ( a+ b \...
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1answer
79 views

Prove $Σ_{cyc}\frac{1}{\left(a+1\right)^2}+\frac{1}{a+b+c+1}\ge 1$

Let $a,b,c>0$ such that $abc=1$. Prove the inequality $$\frac{1}{\left(a+1\right)^2}+\frac{1}{\left(b+1\right)^2}+\frac{1}{\left(c+1\right)^2}+\frac{1}{a+b+c+1}\ge 1$$ I proved $\frac{1}{\left(a+1\...
1
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1answer
53 views

Find the minimum of a three variate function

We consider the function $$f(x,y,z)=(7(x^2+y^2+z^2)+6(xy+yz+zx))(x^2y^2+y^2z^2+z^2x^2)$$ Find $$m=\min\{f(x,y,z):xyz=1\}$$ Using the AM-GM inequality it is clear that $$m_+=\min\{f(x,y,z):xyz=1,x,y,z&...
1
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1answer
94 views

Triangle inequality with the exradii $r_{a}$, $r_{b}$, $r_{c}$, the medians $m_{a}$, $m_{b}$, $m_{c}$

Given a triangle with the exradii $r_{a}$, $r_{b}$, $r_{c}$, the medians $m_{a}$, $m_{b}$, $m_{c}$. Show that $$r_{a}^{2}+ r_{b}^{2}+ r_{c}^{2}\geq 3\sqrt{3}. S+ \left ( m_{a}- m_{b} \right )^{2}+ \...
1
vote
2answers
313 views

Prove $\sum\limits_{cyc}\sqrt[3]{\frac{a^2+bc}{b+c}}\ge\sqrt[3]{9(a+b+c)}$

Let $a,b,c$ be positive real numbers. Show that $$\sum_{cyc}\sqrt[3]{\dfrac{a^2+bc}{b+c}}\ge\sqrt[3]{9(a+b+c)}$$ I have tried C-S and Holder inequalities, without success. How to solve it?
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1answer
87 views

Inequality with $ab+bc+ca=3$ [closed]

Let $a$, $b$, $c>0$ and $ab+bc+ca=3$, prove that $\sum_{cyc}{\frac{a^2+b^2}{a+b+1}}\geq \frac{6abc+6}{a+b+c+3}$. I have tried using Holder $\sum_{cyc}{\frac{a^2+b^2}{a+b+1}}\geq \frac{2(a+b+c)^2}{\...
2
votes
2answers
86 views

How to solve (in)equality of three variables with trigonometric solutions

I'm working through a set of inequality problems and I'm stuck on the following question: Find all sets of solutions for which $$(a^2+b^2+c^2)^2=3(a^3b+b^3c+c^3a)$$ holds. Note that $a,b,c\in\...
3
votes
3answers
215 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
147 views

maximum and minimum of $\frac{a^4+b^4+c^4}{(a+b+c)^4},$

Finding maximum and minimum of $\displaystyle \frac{a^4+b^4+c^4}{(a+b+c)^4},$ Where $a,b,c>0$ and $(a+b+c)^3=32abc$ $\bf{My Attempt}$ with AM-GM inequality $\displaystyle \frac{a+b+c}{3}\geq (abc)^...
1
vote
3answers
208 views

If $a+b+c=0$ prove that $ 3(a^2+b^2+c^2) \times (a^5+b^5+c^5) = 5(a^3+b^3+c^3) \times (a^4+b^4+c^4) $

If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove $$ 3(a^2+b^2+c^2) \times (a^5+b^5+c^5) = 5(a^3+b^3+c^3) \times (a^4+b^4+c^4) $$ I made this question as a more difficult (higher degree) version of ...
0
votes
2answers
66 views

Find the minimum value of $ a/(1+b) + b/(1+a) + (1-a)(1-b)$ [closed]

Find the minimum value of $$ \frac{a}{1+b} + \frac{b}{1+a} + (1-a)(1-b)$$ when $a,b$ is real numbers and satisfying $a,b \in(0,1)$.
2
votes
1answer
104 views

Prove this inequality $3(abc+4)\ge 5(ab+bc+ca)$

Let $\{a,b,c\}\subset\mathbb R$ such that $a+b+c=3$ and $abc\ge -4$. Prove that: $$3(abc+4)\ge 5(ab+bc+ca).$$ *) $ab+bc+ca<0$ This ineq is right *) $ab+bc+ca\ge 0$ then in $ab,bc, ca$ at least a ...
1
vote
3answers
87 views

Is it possible to prove that for $x,y,z \in \Bbb{R}$; if $x,y,z >0$ and $xyz=1$, then $(x+y+z)\ge 3$ without using the AM-GM inequality?

I'm asked in an exercise from an algebra textbook to prove that for$\{ x,y,z\} \subset \Bbb{R}$; if $x,y,z >0$ and $xyz=1$, then $x+y+z\ge 3$. Using the arithmetic and geometric mean inequality ...
5
votes
2answers
329 views

How to prove specific inequality, assuming $\prod\limits_{i=1}^{n}(a_{i}-1)=1$

Let $n\ge 2$ be postive integer and $a_{i},(i=1,2,\cdots,n)$ be real numbers such that $a_{i}>1,(i=1,2,\cdots,n),\prod_{i=1}^{n}(a_{i}-1)=1$. Show that $$\sum_{i=1}^{n}\dfrac{1}{\displaystyle\...
3
votes
1answer
236 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 ...
2
votes
4answers
2k views

Inequality : $\sqrt[3]{\frac{x^3+y^3+z^3}{xyz}} + \sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}} \geq 1+\sqrt[3]{3}$

Let $x>0$, $y>0$ and $z > 0$. Prove that $$\sqrt[3]{\frac{x^3+y^3+z^3}{xyz}} + \sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}} \geq 1+\sqrt[3]{3}.$$ From Micheal Rozenberg's answer : $(x+2)\sqrt{x^2+2}\...
1
vote
1answer
87 views

Inequality $(xy+z)^2+(yz+x)^2+(zx+y)^2\geq\sqrt{2}(x+y)(y+z)(z+x)$

Let $x,y,z\geq 0$. Prove that $$(xy+z)^2+(yz+x)^2+(zx+y)^2\geq\sqrt{2}(x+y)(y+z)(z+x).$$ Expanding gives $$\sum x^2y^2+\sum x^2+6xyz\geq \sqrt{2}\sum x^2y+2\sqrt{2}xyz.$$ If we use AM-GM on the left-...
1
vote
2answers
82 views

Inequality - Maximize k such that

I'm not sure how to solve this problem I came across online. Here it is: Let $a, b, c$ be nonnegative reals. Maximize $k$ such that $$(a+b+c)\left(\dfrac 1a + \dfrac 1b + \dfrac 1c\right) + \...
0
votes
2answers
117 views

How prove this in equality with $abc=1$

let $a>$, $b>0$ and $c>0$ such $abc=1$. Show that: $$(ab+bc+ca)\left(\dfrac{1}{a^2+2}+\dfrac{1}{b^2+2}+\dfrac{1}{c^2+2}\right)\ge 3.$$ It see use $a=\dfrac{x}{y},b=\dfrac{y}{z},c=\dfrac{z}{...
1
vote
2answers
423 views

Maximum value of $a^3 + b^3 + c^3$

If $a+b+c=5$ and $ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} =\frac{1}{5}$ then find the maximum value of $a^3 + b^3 +c^3$ where $a,b,c $ is real numbers. My Attempt writing $a+b=5-c$ and $ \frac{1}{a} +\...
1
vote
1answer
103 views

Inequality $ \sum_{cyc}\frac{b\sqrt{c}}{a(\sqrt{3c}+\sqrt{ab})}\geq \frac{3\sqrt{3}}{4} $

Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=1$. Prove that $$ \sum_{cyc}\frac{b\sqrt{c}}{a(\sqrt{3c}+\sqrt{ab})}\geq \frac{3\sqrt{3}}{4} .$$ My attempt : Substitute $a=x^2, b=y^2, ...