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Questions tagged [buffalo-way]

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

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Help to prove $\frac{a}{(b^2+3)^2}+\frac{b}{(c^2+3)^2}+\frac{c}{(a^2+3)^2}\ge\frac{3}{16}$ for $a+b+c=3.$

If $a,b,c\ge 0: a+b+c=3$ then prove$$\frac{a}{(b^2+3)^2}+\frac{b}{(c^2+3)^2}+\frac{c}{(a^2+3)^2}\ge\frac{3}{16}.$$ I've tried to use Holder inequality as$$\sum_{cyc}a\cdot\sum_{cyc}\frac{a}{(b^2+3)^2}...
Anonymous's user avatar
0 votes
1 answer
89 views

Determining best constant $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+3k\ge (k+1)\cdot\frac{x+y+z}{\sqrt[3]{xyz}}.$

Findind the best constant $k$ satisfying the following inequality$$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+3k\ge (k+1)\cdot\frac{x+y+z}{\sqrt[3]{xyz}},$$holds for all $x,y,z>0.$ It was in AOPS. See ...
Anonymous's user avatar
0 votes
1 answer
88 views

Find proof for sum of cyclic square root inequality where $a+b+c=3.$

While solving one inequality which posted by @arqady-AOPS. (See unsolved inequality.) $$\color{black}{\sqrt{24a^2b+25}+\sqrt{24b^2c+25}+\sqrt{24c^2a+25}\le 21,}$$I arrived at a much simpler, but still ...
Anonymous's user avatar
0 votes
2 answers
92 views

Help to prove $\frac{a^3b+b^3c+c^3a}{a+b+c}+1\ge 2\sqrt{\frac{a^2b^2+b^2c^2+c^2a^2}{a+b+c}}$.

I deal with a problem belonged RMM magazine without success. It's Let $a,b,c>0: abc=1$. Prove that: $$\frac{a^3b+b^3c+c^3a}{a+b+c}+1\ge 2\sqrt{\frac{a^2b^2+b^2c^2+c^2a^2}{a+b+c}}$$ My idea is ...
Sickness's user avatar
1 vote
2 answers
135 views

Find a nice proof for an elegant problem.

Question. For any $a,b,c>0$ then prove that$$\frac{\sqrt{5a^2+4bc}}{bc}+\frac{\sqrt{5b^2+4ca}}{ca}+\frac{\sqrt{5c^2+4ab}}{ab}\le \frac{9}{4}\left(\frac{a^2+b^2+c^2}{abc}+\frac{a+b+c}{ab+bc+ca}\...
Dragon boy's user avatar
0 votes
3 answers
194 views

Prove that $\sum\limits_{\mathrm{cyc}}\sqrt{\frac{xy}{z}}+6\sqrt{xyz}\ge \sum\limits_{\mathrm{cyc}} \sqrt{3x^3+6xyz}$

Context. I saw a problem on facebook.It's Let positive real numbers $x,y,z$ such that $x+y+z=x^2+y^2+z^2$. Prove that$$\sqrt{\frac{xy}{z}}+\sqrt{\frac{yz}{x}}+\sqrt{\frac{zx}{y}}+6\sqrt{xyz}\ge \sqrt{...
Dragon boy's user avatar
3 votes
2 answers
103 views

How to prove $\frac{1}{\sqrt{3a^2+13b}}+\frac{1}{\sqrt{3b^2+13c}}+\frac{1}{\sqrt{3c^2+13a}} \ge \frac{3}{4}$?

Let $a,b,c\ge 0: ab+bc+ca>0$ and $a+b+c=3.$ Prove that $$\frac{1}{\sqrt{3a^2+13b}}+\frac{1}{\sqrt{3b^2+13c}}+\frac{1}{\sqrt{3c^2+13a}} \ge \frac{3}{4}. \tag{*}$$ Here is what I've done so far. By ...
Sickness's user avatar
1 vote
1 answer
188 views

Prove $\sqrt{a+4b}+\sqrt{b+4c}+\sqrt{c+4a}\le 3\sqrt{abc+4}$ when $ a^2+b^2+c^2=3.$

Let $a,b,c\ge 0: a^2+b^2+c^2=3.$ Prove that$$\color{black}{\sqrt{a+4b}+\sqrt{b+4c}+\sqrt{c+4a}\le 3\sqrt{abc+4}. }$$ Naturally, I use Cauchy-Schwarz as $$\sqrt{a+4b}+\sqrt{b+4c}+\sqrt{c+4a}\le \sqrt{...
Dragon boy's user avatar
1 vote
1 answer
145 views

How to prove $\frac{1}{\sqrt{14a^2+b^2+c^2}}+\frac{1}{\sqrt{14b^2+a^2+c^2}}+\frac{1}{\sqrt{14c^2+b^2+a^2}}\ge \frac{9}{4(a+b+c)}$?

Prove that$$\frac{1}{\sqrt{14a^2+b^2+c^2}}+\frac{1}{\sqrt{14b^2+a^2+c^2}}+\frac{1}{\sqrt{14c^2+b^2+a^2}}\ge \frac{9}{4(a+b+c)},$$holds for all $a,b,c>0.$ I tried to use C-S $\dfrac{1}{x}+\dfrac{1}{...
Sickness's user avatar
0 votes
3 answers
140 views

Finding $\small{\min\limits_{a+b+c=2}\sqrt{a-2bc+3}+\sqrt{b-2ca+3}+\sqrt{c-2ab+3}.}$

Given non-negative real numbers $a,b,c$ satisfying $a+b+c=2.$ Find the minimal value of expression $$P=\sqrt{a-2bc+3}+\sqrt{b-2ca+3}+\sqrt{c-2ab+3}.$$ Source: AOPS-Jokehim. See also here. Here is ...
Dragon boy's user avatar
-1 votes
1 answer
97 views

If $a^2+b^2+c^2=3,$ then prove $\sqrt{\frac{a+b}{4a+3}}+\sqrt{\frac{b+c}{4b+3}}+\sqrt{\frac{c+a}{4c+3}}\le 3\sqrt{\frac{2}{7}}.$ [closed]

Question. Given non-negative real numbers $a,b,c$ such that $a^2+b^2+c^2=3.$ Prove that $$\sqrt{\frac{a+b}{4a+3}}+\sqrt{\frac{b+c}{4b+3}}+\sqrt{\frac{c+a}{4c+3}}\le 3\sqrt{\frac{2}{7}}.$$ I tried to ...
Anonymous's user avatar
3 votes
2 answers
118 views

Prove $\sqrt{2a+b^3}+\sqrt{2b+c^3}+\sqrt{2c+a^3}\le 3\sqrt{3}$ for $a^2+b^2+c^2=3.$

Problem. Given non-negative real numbers $a,b,c$ satisfying $a^2+b^2+c^2=3.$ Prove that $$\sqrt{2a+b^3}+\sqrt{2b+c^3}+\sqrt{2c+a^3}\le 3\sqrt{3}.$$ Naturally by using Cauchy-Schwarz, we need to show ...
TATA box's user avatar
1 vote
4 answers
85 views

Help to prove $\frac{1}{4a^2-8ac+13c^2}+\frac{1}{4b^2-8bc+13c^2}\ge \frac{2}{(a+b+c)^2}$

If $a\ge b>c\ge 0,$ then prove $$\frac{1}{4a^2-8ac+13c^2}+\frac{1}{4b^2-8bc+13c^2}\ge \frac{2}{(a+b+c)^2}.$$ When does equality hold ? I tried to use Cauchy-Schwarz $$(a+b+c)^2\geq 2(a^2+b^2)+13c^2-...
Anonymous's user avatar
0 votes
7 answers
141 views

Finding $\small{\max\limits_{a+b+c=3}\frac{1}{\sqrt{a^2-3a+3}}+\frac{1}{\sqrt{b^2-3b+3}}+\frac{1}{\sqrt{c^2-3c+3}}.}$

Let $a,b,c\ge 0: a+b+c=3.$ Find maximum $$P=\frac{1}{\sqrt{a^2-3a+3}}+\frac{1}{\sqrt{b^2-3b+3}}+\frac{1}{\sqrt{c^2-3c+3}}.$$ By $a=b=c=1,$ we got that $P=3$ and we will prove it is desired maximal ...
Dragon boy's user avatar
0 votes
1 answer
49 views

How to find minimum $T= a+b+(a-b)(c-b)$ for $ab+bc+ca=3.$

Let $a \ge b\ge c\ge 0$ such that $ab+bc+ca=3.$ Find minimum $$T= a+b+(a-b)(c-b).$$ For $a=b=c=1,$ T get minimal value is equal to $2.$ Hence, I tried to prove $a+b+(a-b)(c-b)\ge 2.$ My ugly proof ...
Dragon boy's user avatar
2 votes
2 answers
91 views

Prove $\sum\frac{a}{(2k^2-3k+2)a+k^2b+c}\le \frac{1}{k^2-k+1}$

Question If $a,b,c\ge 0:a+b+c>0$ and non-negative constant $k$ then prove$$\frac{a}{(2k^2-3k+2)a+k^2b+c}+\frac{b}{(2k^2-3k+2)b+k^2c+a}+\frac{c}{(2k^2-3k+2)c+k^2a+b}\le \frac{1}{k^2-k+1}$$ Equality ...
Dragon boy's user avatar
3 votes
4 answers
255 views

If $a,b,c\ge0:a+b+c=3,$ prove $\sqrt{a+b+b^2}+\sqrt{b+c+c^2}+\sqrt{c+a+a^2}\ge 3\sqrt{3}.$

Problem. Let $a,b,c\ge0:a+b+c=3.$ Prove that $$\sqrt{a+b+b^2}+\sqrt{b+c+c^2}+\sqrt{c+a+a^2}\ge 3\sqrt{3}.$$ Equality holds at $(1,1,1);(0,0,3).$ I tried to use some classical inequality but it seems ...
TATA box's user avatar
2 votes
2 answers
181 views

How To Prove $a^3b+b^3c+c^3a>a^2b^2+b^2c^2+a^2c^2 $ if $ a > b > c > 0\,$?

How to prove this : $$a^3b+b^3c+c^3a>a^2b^2+b^2c^2+a^2c^2 $$ if we know: $$ a > b > c > 0 $$ My attempt: $$\frac {a^3b+b^3a}{2}>a^2b^2 ...(1)$$ $$\frac {b^3c+c^3b}{2}>c^2b^2...(...
Moustapha_M_I's user avatar
1 vote
1 answer
137 views

Which Buffalo Way is stronger?

Let $f(x, y, z)$ be a homogeneous polynomial and we want to prove $_{cyc}\Sigma f\ge0$. There are two different strat with the same name Buffalo Way for this problem. Let $\min(x, y, z)=x$, expand $_{...
didgogns's user avatar
  • 3,597
1 vote
0 answers
62 views

Smallest positive $k$ such that $\sum_{cyc}\left(\frac{(k-1)(z+x)}{y}-\frac{3}{2}\right)(x-y)^2\geq 0$

Let $x,y,z$ be positive real numbers. Find the smallest positive $k$ such that $$\sum_{cyc}\left(\frac{(k-1)(z+x)}{y}-\frac{3}{2}\right)(x-y)^2\geq 0$$ holds true for $x,y,z$. I proved it for $k=\frac{...
maiar's user avatar
  • 45
-2 votes
1 answer
108 views

About the conjecture $\frac{a^{4}}{133a^{3}+81b^{3}}+\frac{b^{3}}{133b^{3}+81c^{3}}+\frac{c^{4}}{133c^{3}+81a^{3}}-\frac{(a+\sqrt{b}+c)}{214}\ge 0$

Hi working on Prove that $\sum_{\mathrm{cyc}} \frac{214x^4}{133x^3 + 81y^3} \ge x + y + z$ for $x, y, z > 0$ I found the following conjecture : Let $0<a\leq b\leq1< 7/5\leq c $ then it seems ...
Miss and Mister cassoulet char's user avatar
0 votes
0 answers
127 views

Conjecture about : $\sum_{cyc}\frac{214x^4}{133x^3 + 81y^3}\ge\frac{x+y+z}{12}+\sum_{cyc}\frac{13x^4}{8x^3+5y^3}-0.25(x-\frac{2yx}{2y+x})\ge x+y+z$

Hi it's a conjecture refinement in the simple case of Prove that $\sum_{\mathrm{cyc}} \frac{214x^4}{133x^3 + 81y^3} \ge x + y + z$ for $x, y, z > 0$ . Conjecture : Let $x,y,z>0$ such that $x\...
Miss and Mister cassoulet char's user avatar
1 vote
1 answer
87 views

General case of :$\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+d}+\frac{d}{1+a}\leq \frac{a+b+c+d}{1+\frac{1}{4}(a+c)(b+d)}$

Here in my answer (Prove that $\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+d}+\frac{d}{1+a}\le2$ for $0 \le a, b, c, d \le 1$) I show the inequality : Let $0\leq a,b,c,d\leq 1$: $$\frac{a}{1+b}+\frac{b}{1+...
Miss and Mister cassoulet char's user avatar
-1 votes
1 answer
92 views

Generalization of a cyclic inequality : $ \sum_{cyc}{\sqrt{\frac{x+1}{4x^2+10x+4}}}\leqslant 1$

Hi it's a follow up of Prove $\sum_{cyc}{\sqrt{\frac{x+1}{x^2+16x+1}}}\geqslant 1$ and $ \sum_{cyc}{\sqrt{\frac{x+1}{4x^2+10x+4}}}\leqslant 1$ for $x,y,z>0,xyz=1$ : Problem : Let $x_i>0$ such ...
Miss and Mister cassoulet char's user avatar
1 vote
1 answer
93 views

Prove inequality using a given inequality

Question: It is given that $$a^r(a-b)(a-c)+b^r(b-c)(b-a)+c^r(c-a)(c-b)\geq 0$$ for positive integers $a,b,c,r$. We are supposed to show that $$\frac{1}{a^5}+\frac{1}{b^5}+\frac{1}{c^5}+\frac{a+b+c}{a^...
KHOOS's user avatar
  • 409
5 votes
1 answer
166 views

How to prove $\frac{ab^2}{1+2b^2+c^2}+\frac{bc^2}{1+2c^2+a^2}+\frac{ca^2}{1+2a^2+b^2} \le \frac{3}{4}$ if $a+b+c=3$

$a,b,c\ge 0,a+b+c=3.$ Prove: $$\frac{ab^2}{1+2b^2+c^2}+\frac{bc^2}{1+2c^2+a^2}+\frac{ca^2}{1+2a^2+b^2} \le \frac{3}{4}$$ This problem was found in this post. As you can see, no one in that post gave ...
Modern_Hunter's user avatar
1 vote
2 answers
77 views

For any positive real numbers $a,b,c$, show that $\min\{(b-c)^2,(c-a)^2,(a-b)^2)\} \leq \frac{a^2+b^2+c^2}{5}$

For any positive real numbers $a,b,c$, show that $\min\{(b-c)^2,(c-a)^2,(a-b)^2)\} \leq \frac{a^2+b^2+c^2}{5}$ I managed to show this with a 3 instead of a 5 : The inequality is symmetric in $a$, $b$,...
cheubriz's user avatar
0 votes
0 answers
97 views

How to show that the constraint $(C)$ exists?

Let $\sqrt{3}<p\leq x\leq q\leq 8\sqrt{3}$ and $\sqrt{3}< a\leq 8\sqrt{3}$ and the constraint $(C)$ then we have : $$f(x)+h(x)+g(x)\leq \left(\frac{f(p)-f(q)}{p-q}+\frac{g(p)-g(q)}{p-q}\right)(...
Miss and Mister cassoulet char's user avatar
6 votes
0 answers
493 views

$\sqrt{\frac{a^2+4bc}{b^2+c^2}}+\sqrt{\frac{b^2+4ac}{a^2+c^2}}+\sqrt{\frac{c^2+4ba}{b^2+a^2}}\ge 2+\sqrt{2}$ [closed]

Prove that $\forall a,b,c\ge 0$ then $$\sqrt{\frac{a^2+4bc}{b^2+c^2}}+\sqrt{\frac{b^2+4ac}{a^2+c^2}}+\sqrt{\frac{c^2+4ba}{b^2+a^2}}\ge 2+\sqrt{2}$$ I have some ideas but they didn't lead to simpler ...
Frog WeII's user avatar
  • 424
5 votes
1 answer
407 views

A travelled inequality found by discriminant

Given three real numbers $x, y, z$ so that $1\leq x, y, z\leq 8.$ Prove that $$\sum\limits_{cyc}\frac{x}{y}\geq\sum\limits_{cyc}\frac{2x}{y+ z}$$ I found this inequality by discriminant, we realised ...
user avatar
0 votes
1 answer
28 views

Disclude a number in a sequence so that the greatest divisor common of all the others at most. Find that GCD and the position of that discluded one.

Problem. Disclude a number in a sequence so that the greatest divisor common of all the others at most. By programming, find that GCD and the position of that discluded one. My code in the ...
user avatar
5 votes
1 answer
292 views

Cyclic inequality with $a+b+c=1$

Let $a;b;c$ be positive real numbers such that $a+b+c=1$ Prove: $$a^2+b^2+c^2+a^2b+b^2c+c^2a+4abc \geq \dfrac{16}{27}$$ First, I tried with $abc \geq (a+b-c)(b+c-a)(c+a-b)$ But it gave nothing. Then ...
Nguyenthuyhangaz's user avatar
1 vote
1 answer
499 views

Minimizing number of multiplication signs in Computational linear algebra for kind of such these examples

"The main purpose of this note is pedagogical." i.e. $$a^{2}+ b^{2}+ c^{2}- ab- bc- ca=\left ( c- a \right )^{2}- \left ( a- b \right )\left ( b- c \right )$$ $$b^{2}- 4ac=\left ( c- a \...
user avatar
3 votes
0 answers
95 views

Prove $\sum\limits_{cyc} \frac{(a+14)^2}{(b+4)^2}\geq \frac{3}{2} \sum\limits_{cyc} \left(\frac{a+14}{b+4}+\frac{b+4}{a+14}\right)-6$

Problem. For $a,b,c\geq 0.$ Prove that: $$\sum\limits_{cyc} \frac{(a+14)^2}{(b+4)^2}\geq \frac{3}{2} \sum\limits_{cyc} \left(\frac{a+14}{b+4}+\frac{b+4}{a+14}\right)-6$$ I have a solution but it's not ...
NKellira's user avatar
  • 2,001
5 votes
2 answers
344 views

Prove $\sqrt{a + ab} + \sqrt{b} + \sqrt{c} \ge 3$ for $c = \min(a, b, c)$ and $ab + bc + ca = 2$

Problem 1: Let $a, b, c \ge 0$ with $c = \min(a, b, c)$ and $ab + bc + ca = 2$. Prove that $\sqrt{a + ab} + \sqrt{b} + \sqrt{c} \ge 3$. Background: This problem was proposed by csav10@AoPS (https://...
River Li's user avatar
  • 39k
0 votes
0 answers
99 views

Prove $\sum_{\mathrm{cyc}} \sqrt{34x^2 + 28y^2 + 7z^2 - xy - 28yz + 41zx} \ge 9x + 9y + 9z$

Problem 1: Let $x, y, z \ge 0$. Prove that $$\sum_{\mathrm{cyc}} \sqrt{34x^2 + 28y^2 + 7z^2 - xy - 28yz + 41zx} \ge 9x + 9y + 9z. \tag{1}$$ Background: I came up with the problem when I tried to ...
River Li's user avatar
  • 39k
2 votes
2 answers
144 views

Prove $5\Big(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\Big)\geq \frac{a^2+b^2+c^2}{ab+bc+ca}+10.$

Problem. (?) For $a,b,c$ be non-negative numbers such as $a \geq 2(b+c).$ Prove:$$5\Big(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\Big)\geq \frac{a^2+b^2+c^2}{ab+bc+ca}+10.$$ My Solution. We write the ...
NKellira's user avatar
  • 2,001
8 votes
3 answers
800 views

$\sqrt{a^2+5b^2}+\sqrt{b^2+5c^2}+\sqrt{c^2+5a^2}\geq\sqrt{10(a^2+b^2+c^2)+8(ab+ac+bc)}$ for any real numbers.

I think that this inequality is strong, though I do not have knowledge of many techniques. There goes my work: Positive variables only make the inequality stronger, hence suppose $a,b,c\geqslant0$ $$ \...
Book Of Flames's user avatar
1 vote
0 answers
60 views

Intermediate inequality for an inequality from Crux mathematicorum

I propose a little refinement of this an difficult inequality from Crux mathematicorum . Claim: Let $a,b,c>0$ such that $a\geq b\geq c$ and $\frac{a}{c}\geq\frac{b}{a}\geq \frac{c}{b}$ and $b\geq ...
Miss and Mister cassoulet char's user avatar
4 votes
2 answers
146 views

Proving $4\Big(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \Big)+\frac{81}{(a+b+c)^2}\geqslant{\frac {7(a+b+c)}{abc}}$

For $a,b,c>0.$ Prove$:$ $$4\Big(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \Big)+\dfrac{81}{(a+b+c)^2}\geqslant{\dfrac {7(a+b+c)}{abc}}$$ My proof is using SOS$:$ $${c}^{2}{a}^{2} {b}^{2}\Big( \...
NKellira's user avatar
  • 2,001
3 votes
1 answer
196 views

Proving ${\frac {35{x}^{2}+7x(y+z)+23yz}{35(x^2+y^2+z^2)+37(xy+yz+zx)}}\leqslant \sqrt {{\frac {{x}^{2}+yz}{6\,{y}^{2} +6\,yz+6\,{z}^{2}}}}$

For $x,y,z \geqslant 0.$ Proving$:$ $${\dfrac {35{x}^{2}+7x(y+z)+23yz}{35(x^2+y^2+z^2)+37(xy+yz+zx)}}\leqslant \sqrt {{\dfrac {{x}^{2}+yz}{6{y}^{2} +6yz+6{z}^{2}}}}\quad \quad(\text{tthnew})$$ My ...
NKellira's user avatar
  • 2,001
4 votes
5 answers
143 views

Prove that $\frac{1}{abc}+36\ge \frac{21}{ab+bc+ca}$

If $a+b+c=1$ and $a,b,c>0$ then prove that $$\frac{1}{abc}+36\ge \frac{21}{ab+bc+ca}$$ My try : We have to prove: $$ab+bc+ca+36abc(ab+bc+ca)\ge 21abc$$ or after homogenising we get : $$\sum a^4b+3\...
Albus Dumbledore's user avatar
2 votes
4 answers
132 views

Proving $\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$

For $a,b,c\ge 0$ Prove that $$\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$$ My Attempt WLOG $b=\text{mid} \{a,b,c\},$ $$\left(\dfrac{a^3+b^3+c^3}{3}-abc\right)^2-\dfrac{9}{16}(...
BestChoice123's user avatar
0 votes
1 answer
304 views

Proving $3(a^4 + b^4 + c^4 + d^4) + 4abcd \geq (a+b+c+d)(a^3 + b^3 + c^3 + d^3)$

Let $a,b,c,d>0$. Prove that$:$ $$3(a^4 + b^4 + c^4 + d^4) + 4abcd \geqslant (\,a+b+c+d\,)(\,a^3 + b^3 + c^3 + d^{3}\,)$$ As pointed out by @tthnew, this question was posted in: https://...
Trần Hùng Minh's user avatar
0 votes
3 answers
100 views

Proving $\displaystyle \sum_{cyc}\frac{(a^2+b^2)}{a+b}\leqslant \frac{3(a^2+b^2+c^2)}{a+b+c}$ [duplicate]

For $a,b,c>0$, prove $\displaystyle \sum_{cyc}\frac{(a^2+b^2)}{a+b}\leqslant \frac{3(a^2+b^2+c^2)}{a+b+c}$ I've simplified the inequality by multiplying both sides with $(a+b+c).$ So the inequality ...
somerndguy's user avatar
2 votes
1 answer
175 views

Proving $\frac {a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \geqslant \frac 32 \cdot \sqrt[6]{\frac{ab+bc+ca}{a^2+b^2+c^2}}$

For $a,b,c>0.$ Prove$:$ $$\displaystyle \frac {a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \geqslant \frac 32 \cdot \sqrt[6]{\dfrac{ab+bc+ca}{a^2+b^2+c^2}}$$ My try. The Buffalo Way method help here$,$ but ...
NKellira's user avatar
  • 2,001
5 votes
1 answer
80 views

on possible generalisations of $1\le\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\le 2$

Here is a known double inequality for positive numbers: $$ 1\le\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\le 2 $$ Source: https://www.cut-the-knot.org/m/Algebra/Crux4196.shtml I'm curious if at least ...
user807920's user avatar
1 vote
1 answer
100 views

Proving a non-homogeneous inequality with $x,y,z>0$

For $x,y,z>0.$ Prove: $$\frac{1}{2}+\frac{1}{2}{r}^{2}+\frac{1}{3}\,{p}^{2}+\frac{2}{3}\,{q}^{2}-\frac{1}{6} Q-\frac{3}{2} r-\frac{2}{3}q-\frac{1}{6}pq-\frac{5}{3} \,pr\geqslant 0$$ where $$\Big[p=...
NKellira's user avatar
  • 2,001
7 votes
2 answers
172 views

Find maximum $k \in \mathbb{R}^{+}$ such that $ \frac{a^3}{(b-c)^2} + \frac{b^3}{(c-a)^2} + \frac{c^3}{(a-b)^2} \geq k (a+b+c) $

Find maximum $k \in \mathbb{R}^{+}$ such that $$ \frac{a^3}{(b-c)^2} + \frac{b^3}{(c-a)^2} + \frac{c^3}{(a-b)^2} \geq k (a+b+c) $$ for all $a, b, c$ that are distinct positive real numbers ( $a \neq ...
CuriousMind's user avatar
  • 1,590
7 votes
5 answers
796 views

Prove that $\sum_{\mathrm{cyc}} \frac{214x^4}{133x^3 + 81y^3} \ge x + y + z$ for $x, y, z > 0$

Problem. Let $x, y, z > 0$. Prove that $$\frac{214x^4}{133x^3 + 81y^3} + \frac{214y^4}{133y^3 + 81z^3} + \frac{214z^4}{133z^3 + 81x^3} \ge x+y+z.$$ It is verified by Mathematica. The inequality ...
River Li's user avatar
  • 39k