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Questions tagged [buffalo-way]

Inequalities that can be proved by BW (Buffalo Way).

-2
votes
1answer
39 views

Prove $\sum\limits_{cyc}\frac{ab}{b^{\,2}+ c^{\,2}}\geqq \frac{3}{2}$

For $a\geqq b\geqq c> 0$. Prove $$\frac{ab}{b^{\,2}+ c^{\,2}}+ \frac{bc}{c^{\,2}+ a^{\,2}}+ \frac{ca}{a^{\,2}+ b^{\,2}}\geqq \frac{3}{2}$$ I used discrim to find and I want to see a solution ...
1
vote
2answers
40 views

Prove $2\sum\limits_{cyc}\,a^{\,3}+ 3\,abc\geqq 3\sum\limits_{cyc}\,a^{\,2}b$

For $a,\,b,\,c\geqq 0$ and $b\equiv {\rm mid}\,\{\,a,\,b,\,c\,\}$. Prove $$2\sum\limits_{cyc}\,a^{\,3}+ 3\,abc\geqq 3\sum\limits_{cyc}\,a^{\,2}b$$ Inspried from $\lceil$ Prove $k=0$ is the only non-...
-3
votes
1answer
46 views

Prove $\sum\limits_{cyc}\,\frac{a^{\,2}+ k\,b^{\,2}}{k\,a^{\,2}+ b^{\,2}}\geqq 3$

For $a\geqq b\geqq c> 0$. Prove $$\sum\limits_{cyc}\,\frac{a^{\,2}+ k\,b^{\,2}}{k\,a^{\,2}+ b^{\,2}}\geqq 3 \tag{SHED}$$ with $k= \frac{b}{c}\geqq 1$. I used SHEDtechniQ to find and I want to ...
0
votes
0answers
57 views

Prove $\sum\limits_{cyc}\,\frac{a}{\sqrt{b(\,a+ b\,)}}\geqq \sum\limits_{cyc}\,\frac{a}{\sqrt{b(\,c+ a\,)}}$ with $a,\,b,\,c> 0$

Let $a,\,b,\,c$ be positive numbers. Prove that $$\sum\limits_{cyc}\,\frac{a}{\sqrt{b(\,a+ b\,)}}\geqq \sum\limits_{cyc}\,\frac{a}{\sqrt{b(\,c+ a\,)}}$$ I tried Holder and $\lceil$ https://...
0
votes
1answer
55 views

Prove $k=0$ is the only non-negative $k$ such that $\sum\limits_{cyc}\,a^{\,3}- \sum\limits_{cyc}\,a^{\,2}b\geqq k(\,a- b\,)(\,a- c\,)(\,b+ c\,)$

Prove with $a+ b,\,b+ c,\,c+ a\geqq 0$ $$\begin{equation}\begin{split} k= constant= 0 \end{split}\end{equation}$$ is the only non-negative $k$ such that $$\begin{equation}\begin{split} \sum\limits_{...
0
votes
1answer
82 views

Prove $\frac{1}{a+ 2\,b}+ \frac{1}{b+ 2\,c}+ \frac{1}{c+ 2\,a}\leqq 1$ with $8\,abc\geqq a+ b+ c+ 5$ and $a,\,b,\,c> 0$

Prove $$\frac{1}{a+ 2\,b}+ \frac{1}{b+ 2\,c}+ \frac{1}{c+ 2\,a}\leqq 1$$ with $8\,abc\geqq a+ b+ c+ 5$ and $a,\,b,\,c> 0$ $$constant= 8$$ is the best $constant$, which was found by me (using ...
0
votes
1answer
215 views

Inequality for $a,b,c>0$ $\sum_{cyc}\sqrt{\frac{a^3}{14a^2+4b^2}}\leq \sum_{cyc}\sqrt{\frac{a+b}{36}}$

A friend gives me the following result : Let $a,b,c>0$ then we have : $$\sqrt{\frac{a^3}{14a^2+4b^2}}+\sqrt{\frac{b^3}{14b^2+4c^2}}+\sqrt{\frac{c^3}{14c^2+4a^2}}\leq \sqrt{\frac{a+b}{36}}+\...
2
votes
1answer
73 views

A hard inequality $(a^2-ab+b^2 )(b^2-bc+c^2 )(c^2-ca+a^2 ) + 11abc \leq 12$

Given $$c=\min⁡(a,b,c)~, \quad a+b+c=3 \\ P=(a^2-ab+b^2 )(b^2-bc+c^2 )(c^2-ca+a^2 )~,$$ I have to prove that $$P+11abc \le 12~.$$ I started with $$b^2-bc+c^2 \le b^2 \quad \text{and} \quad c^2-ca+...
4
votes
1answer
97 views

Proving $ \sum_{cyc}^{} \frac {a(a^3+b^3)}{a^2+ab+b^2} \ge \frac{2}{3} (a^2+b^2+c^2)$ for $a, b, c > 0$

For $a,b,c>0$, I have to prove that $$ \sum_{cyc}^{} \frac {a(a^3+b^3)}{a^2+ab+b^2} \ge \frac{2}{3} (a^2+b^2+c^2).$$ We have: $$\begin{align} \sum_{cyc}^{} \frac {a(a^3+b^3)}{a^2+ab+b^2} &= \...
1
vote
3answers
92 views

Prove that $\sum_{cyc}\frac{a}{b^2}\ge 3\sum_{cyc}\frac{1}{a^2}$ for $a,b,c>0$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$

$a$, $b$ and $c$ are three positives such that $\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = 1$. Prove that $$\dfrac{a}{b^2} + \frac{b}{c^2} + \frac{c}{a^2} \ge 3 \cdot \left(\frac{1}{a^2} + \frac{1}{...
1
vote
1answer
75 views

Prove $F(\,k\,)=(\,16\,X^{\,}- 24\,X+ 18\,)k^{\,2}- 11\,X+ 1\geqq 0$

Given that $1\leqq X\leqq k,$ prove that $$F(\,k\,)=(\,16\,X^{\,}- 24\,X+ 18\,)k^{\,2}- 11\,X+ 1\geqq 0 \tag{29}$$ Origin For $a,\,b,\,c\geqq 0$ and $a+ b+ c= 3$, prove that $(2+a^2)(2+b^2)(2+c^2)+...
1
vote
2answers
87 views

Prove $\frac{a}{b+ c}+ \frac{b}{c+ a}+ \frac{c}{a+ b}+ \frac{63}{5}\left [ \frac{2\,c^{\,2}}{(\,a+ b\,)^{\,2}}- \frac{c}{a+ b} \right ]\geqq 0$

Given that $a,\,b,\,c> 0$, prove: $$\frac{a}{b+ c}+ \frac{b}{c+ a}+ \frac{c}{a+ b}+ \frac{63}{5}\left [ \frac{2\,c^{\,2}}{(\,a+ b\,)^{\,2}}- \frac{c}{a+ b} \right ]\geqq 0$$ I only tried Buffalo ...
1
vote
0answers
148 views

Prove $\frac{x}{y}+ \frac{y}{z}+ \frac{z}{x}\geqq \frac{2\,x}{y+ z}+ \frac{2\,y}{z+ x}+ \frac{2\,z}{x+ y}$ for $x,\,y,\,z\in [\,j,\,k\,j\,]$

For $x,\,y,\,z\in [\,j,\,k\,j\,]$ $$\frac{x}{y}+ \frac{y}{z}+ \frac{z}{x}\geqq \frac{2\,x}{y+ z}+ \frac{2\,y}{z+ x}+ \frac{2\,z}{x+ y}$$ is true with $j= constant,\,k= constant> 8$ and $k_{\,\max}$...
1
vote
1answer
98 views

Find $k=constant$ such that $f(a,\,b,\,c\,)=\frac{3a+2b}{\sqrt{5a^2-ab+b^2}}+\frac{3b+2c}{\sqrt{5b^2-bc+c^2}}\leqq f(a,k\,a+\sqrt[3]{abc}-k\,c,c\,)$

Give $3$ positve numbers $a,\,b,\,c$ such that $abc= 1$ , prove: $$f\left ( a,\,b,\,c \right )= \frac{3\,a+ 2\,b}{\sqrt{5\,a^{\,2}- ab+ b^{\,2}}}+ \frac{3\,b+ 2\,c}{\sqrt{5\,b^{\,2}- bc+ c^{\,2}}}\...
9
votes
1answer
321 views

Prove $3\sum\limits_{cyc}\,a^{\,2}b(\,a- b\,)\geqq b(\,a+ b- c\,)(\,a- c\,)(\,c- b\,)$

If you are interested in IMO 1983 We have $$\sum\limits_{cyc}\,a^{\,2}b(\,a- b\,)= \frac{bc(\,a+ b- c\,)^{\,2}(\,a- b\,)^{\,2}}{b(\,a+ b- c\,)+ 3\,a(\,c+ a- b\,)}+ $...
5
votes
1answer
126 views

to prove this inequality $\sum a^3b+3\ge 2(ab+bc+ca)$

let $a,b,c>0$ and such $a+b+c=3$,show that $$a^3b+b^3c+c^3a+3\ge 2(ab+bc+ac)$$ This problem is from my question when $n=3$ case,I found not to prove it. show this inequality with $\sum_{i=1}^{n}...
10
votes
2answers
414 views

show this inequality with $\sum_{i=1}^{n}a_{i}=n$

Let $n\ge 3$ be postive number,$a_{i}>0,i=1,2,\cdots,n$,and $\displaystyle\sum_{i=1}^{n}a_{i}=n$,show that $$a^3_{1}a_{2}+a^3_{2}a_{3}+\cdots+a^3_{n}a_{1}+n\ge 2(a_{1}a_{2}\cdots a_{n-1}+a_{2}a_{...
0
votes
1answer
54 views

Little inequality

After deconditioning and calculating some stuff I obtained this inequality, which looks simple but I can't get to group the termens conveniently. $x^3 + y^3 + z^3 + 6(x^2z + y^2x + z^2y) ≥ 12xyz + 3(x^...
2
votes
1answer
97 views

Inequality $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+15\sqrt[3]{abc}\geq 6(a+b+c).$

Let $a,b$ and $c$ are positive numbers. Prove that $$\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}+15\sqrt[3]{abc}\geq 6(a+b+c).$$ I tried BW, SOS-Schur, but without any success. I let $abc=1$. Then ...
7
votes
2answers
156 views

Proving $\sum_{\text{cyc}} \frac{a}{b^2+c^2+d^2} \geq \frac{3\sqrt{3}}{2}\frac{1}{\sqrt{a^2+b^2+c^2+d^2}}$

Prove that $$\frac{a}{b^2+c^2+d^2}+\frac{b}{a^2+c^2+d^2}+\frac{c}{a^2+b^2+d^2}+\frac{d}{a^2+b^2+c^2}≥\frac{3\sqrt{3}}{2}\frac{1}{\sqrt{a^2+b^2+c^2+d^2}}$$ What I tried, was to say that $a^2+b^2+c^...
1
vote
1answer
105 views

Inequality from AMM problems section

This is Problem 12024 of AMM. It asks to show that if $x,y,z$ are positive reals, and $xyz=1$, then $(x^{10}+y^{10}+z^{10})^{2}\geq 3(x^{13}+y^{13}+z^{13})$. I could show it for the particular case ...
2
votes
4answers
131 views

If $0⩽x⩽y⩽z⩽w⩽u$ and $x+y+z+w+u=1$, prove $xw+wz+zy+yu+ux⩽\frac15$

If $0⩽x⩽y⩽z⩽w⩽u$ and $x+y+z+w+u=1$, prove$$ xw+wz+zy+yu+ux⩽\frac15. $$ I have tried using AM-GM, rearrangement, and Cauchy-Schwarz inequalities, but I always end up with squared terms. For example, ...
0
votes
1answer
75 views

Seeking the Maximum of a Product expression using Inequalities

Since $\ln a\leq a-1,$ I tried to use that inequality to calculate a maximum value for $$\prod_{cyc}\frac{9y+4z-6x}{x}$$ where $x,y,z>0$. Or $$\sum_{cyc}\log\frac{9y+4z-6x}{x}$$ Then $$\log\frac{9y+...
-1
votes
1answer
80 views

$(a+b^2)(b+c^2)(c+a^2)≤13+…$

Prove the following inequality for non-negative real numbers $a,b,c$: If $a+b+c=3$ then: $$(a+b^2)(b+c^2)(c+a^2)\le13+abc(1-2abc)\qquad(1)$$ There are two more variants of the same problem. The ...
7
votes
3answers
494 views

prove this inequality by $abc=1$

Let $a,b,c>0$ and $abc=1$,show that $$(a^{10}+b^{10}+c^{10})^2\ge 3(a^{14}+b^{14}+c^{14})$$ since $$LHS=\sum \left(a^{20}+\dfrac{2}{a^{10}}\right)$$ it is prove $$\sum_{cyc}\left(a^{20}+\dfrac{2}{...
0
votes
2answers
109 views

Proving $a^4+b^4+c^4+(\sqrt {3}-1)(a^2 b c+a b^2 c+a b c^2 )\ge \sqrt {3} (a^3 b+b^3 c+c^3 a)$ for real $a$, $b$, $c$

If $a$, $b$, $c$ are real numbers, I have to prove: $$a^4+b^4+c^4+(\sqrt {3}-1)(a^2 b c+a b^2 c+a b c^2 )\ge \sqrt {3} (a^3 b+b^3 c+c^3 a)$$ Since $$a^4+b^4+c^4 \ge abc(a+b+c)$$ then it is enough ...
5
votes
1answer
180 views

Prove that $(x-y)(y-z)(z-x) \leq \frac{1}{\sqrt{2}}$

If $x,y,z$ are real and $x^2+y^2+z^2=1$, prove that$$(x-y)(y-z)(z-x) \leq \frac{1}{\sqrt{2}}.$$
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\...
0
votes
2answers
74 views

Minimum value of rational expression

If $x,y,z$ are distinct real number, Then minimum value of $\displaystyle \left(\frac{x}{y-z}\right)^2+\left(\frac{y}{z-x}\right)^2+\left(\frac{z}{x-y}\right)^2$ Try: $$\left(\frac{x}{y-z}\right)^2+...
3
votes
7answers
608 views

Elementary proof for $\sqrt[3]{xyz} \leq \dfrac{x+y+z}{3}$

I am searching for an elementary proof of the AM-GM inequality in three variables: $\sqrt[3]{xyz} \leq \dfrac{x+y+z}{3}$ The inequality of the geometric mean vs the arithmetic mean of two variables ...
1
vote
2answers
104 views

let $ \ \ 0<a \leq b \leq c \in \mathbb{R}$ then prove that :

let $ \ \ 0<a \leq b \leq c \in \mathbb{R}$ then prove that : $$\frac{(c-a)^2}{6c}\leq \frac{a+b+c}{3}-\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$ I do not know where to start please help ...
6
votes
2answers
266 views

Turkevicius inequality

Let $a$, $b$, $c$ and $d$ be positive real numbers. Prove that: $$a^4+b^4+c^4+d^4+2abcd \geq a^2b^2+ a^2c^2+ a^2d^2+ b^2c^2+ b^2d^2+ c^2d^2.$$ Source : Inequalities, Theorems, Techniques and ...
2
votes
1answer
87 views

Strange inequality over the positive reals

Let $a$, $b$ and $c$ be positive real numbers with $2a^3b+2b^3c+2c^3a=a^2b^2+b^2c^2+c^2a^2$. Prove that $$2ab(a-b)^2+2bc(b-c)^2+2ca(c-a)^2 \geqslant (ab+bc+ca)^2$$
3
votes
5answers
128 views

Question based on finding maximum and minimum value

If $x$, $y$ and $z$ are distinct positive integers and $x+y+z=11$ then what is the maximum value of $(xyz+xy+yz+zx)$? We know that product is maximum when difference between $x$, $y$ and $z$ is ...
0
votes
1answer
140 views

If $a,b,c,d>0$ prove $\frac{2a^2+bc}{a+b+c} +\frac{2b^2+cd}{b+c+d}+\frac{2c^2+da}{c+d+a}+\frac{2d^2+ab}{d+a+b} \geq a+b+c+d$

If $a,b,c,d>0$ prove $$\frac{2a^2+bc}{a+b+c} +\frac{2b^2+cd}{b+c+d}+\frac{2c^2+da}{c+d+a}+\frac{2d^2+ab}{d+a+b} \geq a+b+c+d$$ I tried cauchy: $$\frac{2a^2+bc}{a+b+c} +\frac{2b^2+cd}{b+c+d}+\frac{...
0
votes
4answers
79 views

Prove the following inequality: $a^2+c^2-b^2>\frac{a^2c^2}{b^2}$

Suppose we the following real numbers: {$a$,$b$,$c$} $\in \mathbb{R}^+$, where $a>b>c$. Is there any way to prove the following: $a^2+c^2-b^2>\frac{a^2c^2}{b^2}$
-1
votes
2answers
90 views

Given $a,b,c$. Prove inequality

Given $a,b,c$ are non-negative number. Prove that $$3(a^2+b^2+c^2)\ge\left(a+b+c\right)\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+(a-b)^2+(b-c)^2+(c-a)^2\ge(a+b+c)^2$$ By C-S: $3(a^2+b^2+c^2)\ge(a+b+...
2
votes
1answer
210 views

If $a+b+c=ab+ac+bc$ then $\sum\limits_{cyc}\frac{a}{a^2+2b}\leq1$.

Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=ab+ac+bc$. Prove that: $$\frac{a}{a^2+2b}+\frac{b}{b^2+2c}+\frac{c}{c^2+2a}\leq1$$ We can try the following. Replace $a\rightarrow\frac{1}{...
4
votes
3answers
218 views

How find this maximum of the value $\sum_{i=1}^{6}x_{i}x_{i+1}x_{i+2}x_{i+3}$?

Let $$x_{1},x_{2},x_{3},x_{5},x_{6}\ge 0$$ such that $$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=1$$ Find the maximum of the value of $$\sum_{i=1}^{6}x_{i}\;x_{i+1}\;x_{i+2}\;x_{i+3}$$ where $$x_{7}=x_{1}...
2
votes
1answer
84 views

For what values of $k>0$ does $abc=1 \implies \sum_{\mbox{cyc}}\left(\frac1{a+k}-\frac{a}{a^2+k}\right) \geq 0$?

For what values of $k>0$ does $$abc=1 \implies \left(\frac1{a+k}-\frac{a}{a^2+k}\right) + \left(\frac1{b+k}-\frac{b}{b^2+k}\right)+ \left(\frac1{c+k}-\frac{c}{c^2+k}\right)\geq 0$$ This question ...
2
votes
1answer
433 views

Why is the “Buffalo Way” considered inelegant?

I was going through an "article" on the "Buffalo Way", where the author said that one should NEVER use the Buffalo Way for proving inequalities in actual real-time contests as it is "highly inelengant"...
12
votes
3answers
542 views

Proving $\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$

The inequality: $$\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$$ Conditions: $a,b,c,d \in \mathbb{R^+}$...
4
votes
1answer
394 views

For $abc=1$ prove that $\sum\limits_\text{cyc}\frac{1}{a+3}\geq\sum\limits_\text{cyc}\frac{a}{a^2+3}$

Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that: $$\frac{1}{a+3}+\frac{1}{b+3}+\frac{1}{c+3}\geq\frac{a}{a^2+3}+\frac{b}{b^2+3}+\frac{c}{c^2+3}$$ I tried TL, BW, the Vasc's ...
4
votes
4answers
190 views

Let $a,b,c$ be the length of sides of a triangle then prove that $a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge0$ [duplicate]

Let $a,b,c$ be the length of sides of a triangle then prove that: $a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge0$ Please help me!!!
3
votes
4answers
133 views

Prove that $(a+b+c-d)(a+c+d-b)(a+b+d-c)(b+c+d-a)\le(a+b)(a+d)(c+b)(c+d)$

Let $a,b,c,d>0$. Prove that $$(a+b+c-d)(a+c+d-b)(a+b+d-c)(b+c+d-a)\le(a+b)(a+d)(c+b)(c+d)$$ I don't know how to begin to solve this problem
5
votes
2answers
156 views

Prove that $\left | \frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a} \right | < \frac{1}{8}.$

Let $a,b,$ and $c$ be the lengths of the sides of a triangle. Prove that $$\left | \dfrac{a-b}{a+b}+\dfrac{b-c}{b+c}+\dfrac{c-a}{c+a} \right | < \dfrac{1}{8}.$$ The best idea I had was to expand ...
11
votes
2answers
448 views

cyclic three variable inequality

Let $a,b,c$ be nonnegative real numbers and $a+b+c=3$. Prove the inequality $$ \sqrt{24a^2b+25}+\sqrt{24b^2c+25}+\sqrt{24c^2a+25}\le 21 $$ I have tried to find the solution using classical ...
8
votes
1answer
144 views

Prove this inequality $a^4+b^4+c^4+9\ge 4(a^2b+b^2c+c^2a)$

let $a,b,c>0$ and such $abc=1$, show that $$a^4+b^4+c^4+9\ge 4(a^2b+b^2c+c^2a)$$ Schur inequality $$a^4+b^4+c^4\ge \sum_{cyc}ab(a^2+b^2)-abc(a+b+c)$$ It suffices to show that $$\sum_{cyc}ab(a^2+...
5
votes
2answers
243 views

For any triangle with sides $a$, $b$, $c$, show that $a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 0$

For any triangle with sides $a$, $b$, $c$: $$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 0$$ I tried substituting $a=x+y$, $b=y+z$, $c=z+x$ but well it doesn't help in any sense except wasting 3 pages that ...
4
votes
1answer
158 views

How prove this Nice inequality $\sum\limits_\text{cyc}\frac{x^2}{y}\ge 3+\frac{x^4+y^4+z^4-x^2-y^2-z^2}{x^3+y^3+z^3-xyz}$

Question: let $x,y,z>0$, show that $$\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\ge 3+\dfrac{x^4+y^4+z^4-x^2-y^2-z^2}{x^3+y^3+z^3-xyz}$$ I know this well know inequality $$\dfrac{x^2}{y}+\...