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.

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-4
votes
1answer
142 views

Conjecture-refinement related to Nesbitt's inequality

Prove or disprove : $ a\geq b\geq c\geq 1$: $$\sum_{cyc}\frac{a}{b+c}\geq\sqrt{\frac{9}{4}+\frac{9}{4}\frac{(c^4+a^4+b^4-c^2a^2-b^2a^2-c^2b^2)}{(3(abc)^{\frac{1}{3}})(a+b)(b+c)(c+a)}+\frac{(c^2+a^2+b^...
1
vote
0answers
61 views

About a cyclic expression problem

Let real numbers $a,b,c>0$ such that $a+b+c=1$ and $ab+bc+ca=q$, where $q>0$. We consider the following expression: $$S_1=\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}.$$ Find the minimum (...
1
vote
1answer
71 views

Prove that $\sqrt{a^2-a+1}+\sqrt{b^2-b+1}+\sqrt{c^2-c+1}\ge a+b+c$ [duplicate]

For $a,b,c>0;abc=1.$ Prove that $$\sqrt{a^2-a+1}+\sqrt{b^2-b+1}+\sqrt{c^2-c+1}\ge a+b+c$$ I will post my solution in the answer. Now I'm looking forward to another solution.
9
votes
0answers
216 views

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 ...
0
votes
0answers
198 views

Hard refinement concerning the inequality $\frac{1}{7a+b}+\frac{1}{7b+c}+\frac{1}{7c+a}\leq\frac{\sqrt{3}}{8}$

For me really it's a nightmarish refinement : Claim : Let $a,b,c>0$ such that $abc=a+b+c$ then we have : $$\frac{1}{7a+b}+\frac{1}{7b+c}+\frac{1}{7c+a}\leq \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\...
2
votes
2answers
67 views

Find maximize of $P=\frac{x\sqrt{yz}}{\sqrt{x^2+1}\sqrt[4]{\left(y^2+4\right)\left(z^2+9\right)}}$

Let $x,y,z\in \mathbb{R^+}$ such that $6x+3y+2z=xyz$. Find maximize of $$P=\frac{x\sqrt{yz}}{\sqrt{x^2+1}\sqrt[4]{\left(y^2+4\right)\left(z^2+9\right)}}$$ We will prove $$P\le \sqrt{\frac{16}{27}}$$ $...
1
vote
1answer
99 views

Prove $3\left(9-5\sqrt{3}\right) \sum \frac{1}{a} \geqslant \sum a^2+\frac32\cdot\frac{\left[(\sqrt3-2)(ab+bc+ca)+abc\right]^2}{abc}$

Let $a,\,b,\,c$ are positive real numbers satisfy $a+b+c=3.$ Prove that $$3\left(9-5\sqrt{3}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant a^2+b^2+c^2 + \frac32 \cdot \frac{\left[(\...
5
votes
3answers
146 views

prove that $\sum_{cyc}\frac{a}{b^2+c^2}\ge \frac{4}{5}\sum_{cyc}\frac{1}{b+c}$

prove that $$\sum_{cyc}\frac{a}{b^2+c^2}\ge \frac{4}{5}\sum_{cyc}\frac{1}{b+c}$$ for positives $a,b,c$ Attempt: By C-S; $$\left(\sum_{cyc}\frac{a}{b^2+c^2} \right) \left(\sum_{cyc} a(b^2+c^2) \right)\...
9
votes
3answers
638 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 $a,b,c\geqslant0$ $$ \...
2
votes
4answers
107 views

Minimize $(x+y)(y+z)(z+x)$ given $xyz(x+y+z) = 1$

$x,y,z$ are positive reals and I am given $xyz(x+y+z) = 1$. Need to minimize $(x+y)(y+z)(z+x)$. Here is my approach. Using AM-GM inequality $$ (x+y) \geqslant 2 \sqrt{xy} $$ $$ (y+z) \geqslant 2 \...
4
votes
1answer
126 views

$(ab + bc + ca) \left(\frac {1}{(a + pb)(a + qb)} + \frac {1}{(b + pc)(b + qc)} + \frac{1}{(c + pa)(c + qa)}\right)\ge \frac {9}{(p + 1)(q + 1)}$

One of my friends showed me this inequality. $$(ab + bc + ca) \left(\frac {1}{(a + pb)(a + qb)} + \frac {1}{(b + pc)(b + qc)} + \frac{1}{(c + pa)(c + qa)}\right)\ge \frac {9}{(p + 1)(q + 1)}$$ for ...
0
votes
4answers
116 views

prove that $\sum_{cyc}\frac{a(b^2+c^2)}{a^2+bc}\ge 3$

if $a,b,c$ are positive prove $$\sum_{cyc}\frac{a(b^2+c^2)}{a^2+bc}\ge 3$$ given $ab+bc+ca=3$ My try:using AM-GM and Titu's lemma : $$\sum_{cyc}\frac{a(b^2+c^2)}{a^2+bc}\ge 2\sum_{cyc}\frac{abc}{a^2+...
0
votes
1answer
71 views

Proving $\sum \frac{a+b}{c} \geq 2.\sqrt{(a+b+c)(\frac{a}{bc} +\frac{b}{ca}+ \frac{c}{ab})}$

Problem. (Le Khanh Sy) For $a,b,c>0.$ Prove$:$ $$\sum \dfrac{a+b}{c} \geq 2\sqrt{(a+b+c)\Big(\dfrac{a}{bc} +\dfrac{b}{ca}+ \dfrac{c}{ab}\Big)}$$ My proof. After squaring ... it's $$4\,{b}^{2}{c}^{2}...
4
votes
3answers
111 views

Proving $(a+b+c) \Big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Big) \leqslant 25$

For $a,b,c \in \Big[\dfrac{1}{3},3\Big].$ Prove$:$ $$(a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) \leqslant 25.$$ Assume $a\equiv \text{mid}\{a,b,c\},$ we have$:$ $$25-(a+b+c) \Big(\dfrac{...
2
votes
2answers
147 views

Prove $\sum ab \sum \frac{1}{(a+b)^2} \geqslant \frac{9}{4}+\frac{kabc\sum (a^2-bc)}{(a+b+c)^3(ab+bc+ca)}$ for the best k.

For $a,b,c\geqslant 0;ab+bc+ca>0.$ Find $k_\max$ and proving in that case$:$ $$(ab+bc+ca)\Big(\dfrac{1}{(a+b)^2}+\dfrac{1}{(b+c)^2}+\dfrac{1}{(c+a)^2}\Big) \geqslant \dfrac{9}{4}+\dfrac{kabc(a^2+b^...
4
votes
2answers
159 views

Proving $\frac{a}{b^3}+\frac{b}{c^3}+\frac{c}{a^3}\geqslant \frac{a+b}{b^3+c^3}+\frac{b+c}{c^3+a^3}+\frac{c+a}{a^3+b^3}$

For $a,b,c>0.$ Prove$:$ $$\dfrac{a}{b^3}+\dfrac{b}{c^3}+\dfrac{c}{a^3}\geqslant \dfrac{a+b}{b^3+c^3}+\dfrac{b+c}{c^3+a^3}+\dfrac{c+a}{a^3+b^3}\quad (\text{Tran Quoc Thinh}) $$ It's easy with ...
4
votes
2answers
119 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( \...
2
votes
3answers
89 views

prove that $\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge 3+n$

prove that $$\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge 3+n$$ where $a,b,c>0$ and $n\ge0,n\le 3$,$a+b+c=ab+bc+ca$ My try: by given condition $a+b+c=ab+bc+ca$ we have $a+b+c\le a^2+b^2+c^2$ ...
2
votes
2answers
96 views

Proving $Q=\frac{a^{3}+b^{3}+c^{3}}{(a+b)(b+c)(c+a)}+k\cdot \frac{(ab+bc+ca)}{(a+b+c)^{2}}\geqslant \frac{3}{8}+\frac{k}{3}$

For $a,b,c\geqslant 0.$ Prove: $$\text{Q}=\frac{a^{3}+b^{3}+c^{3}}{(a+b)(b+c)(c+a)}+k\cdot \frac{(ab+bc+ca)}{(a+b+c)^{2}}\geqslant \frac{3}{8}+\frac{k}{3},$$ where $k={\frac {27}{8}}+\frac{9\sqrt{3}}{...
3
votes
1answer
146 views

Proving $\sum \frac{b+c}{9(a^2+3bc)+4(a+b+c)^2}\geqslant \frac{1}{4(a+b+c)}$

For $a,b,c>0.$ Prove$:$ $$\sum \dfrac{b+c}{9(a^2+3bc)+4(a+b+c)^2}\geqslant \dfrac{1}{4(a+b+c)}$$ SOS solution$:$ $$\dfrac{1}{8(a+b+c)}\sum{\dfrac { \left( 52\,{a}^{2}+95\,ab-142\,ac+52\,{b}^{2}-142\...
0
votes
4answers
129 views

Symmetric Inequality for positive real numbers

Show that if $a,b,c$ are positive real numbers such that $S_2 := ab+bc+ca = 3$, then $$ \cfrac{1}{a+b+2} + \cfrac{1}{a+c+2} + \cfrac{1}{b+c+2} \leq \cfrac{13 \cdot S_1 + 27}{16 \cdot S_1 + 40}$$ where ...
3
votes
5answers
118 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\...
6
votes
2answers
128 views

Find the inequality with the best possible $k= constant$ (with the condition $x^{2}+ y^{2}\leq k$).

Find the inequality with the best possible $constant$ Given two non-negative numbers $x, y$ so that $x^{2}+ y^{2}\leq \frac{2}{7}$. Prove that $$\frac{1}{1+ x^{2}}+ \frac{1}{1+ y^{2}}+ \frac{1}{1+ xy}...
4
votes
0answers
134 views

Proving $\sum_{\text{cyc}}\, (23a-5b-c)(a-b)^2(a+b-3c)^2 \geqslant 0$

Problem (KaiRain's problem). For $a,b,c\geqslant 0.$ Prove $$\displaystyle \sum_{\text{cyc}}\, (23a-5b-c)(a-b)^2(a+b-3c)^2 \geqslant 0$$ I only found a proof by $pqr.$ (Note that from pqr's proof we ...
2
votes
4answers
103 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}(...
1
vote
1answer
95 views

Find the stronger version of $9 ( a+b+c ) ^{2} ( ab+ac+bc ) ^{2}+108a^2b^2c^2-31abc ( a+b+c ) ^{3} \geqslant 0$

For $a,b,c \geqslant 0.$ Then $$9 ( a+b+c ) ^{2} ( ab+ac+bc ) ^{2}+108a^2b^2c^2-31abc ( a+b+c ) ^{3} \geqslant 0.$$ I use computer and found that the following stronger inequality holds for all reals ...
1
vote
0answers
88 views

Proving $ \sum \frac{a(b+c)}{a^2+bc}+\frac{2\sum (a^2-bc)}{(a+b+c)^2}+\frac{96(a-b)^2(b-c)^2(c-a)^2}{(a+b+c)^6} \leqslant \frac{(a+b+c)^2}{ab+bc+ca} $

For $a,b,c>0.$ Prove$:$ $$\displaystyle \frac{a(b+c)}{a^2+bc}+\frac{b(c+a)}{b^2+ca}+\frac{c(a+b)}{c^2+ab} +\frac{2(a^2+b^2+c^2-ab-bc-ca)}{(a+b+c)^2}+\frac{96(a-b)^2(b-c)^2(c-a)^2}{(a+b+c)^6} \...
0
votes
3answers
93 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 ...
2
votes
7answers
195 views

Proving $(1+a^2)(1+b^2)(1+c^2)\geq8 $

I tried this question in two ways- Suppose a, b, c are three positive real numbers verifying $ab+bc+ca = 3$. Prove that $$ (1+a^2)(1+b^2)(1+c^2)\geq8 $$ Approach 1: $$\prod_{cyc} {(1+a^2)}= \left({...
1
vote
1answer
138 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 ...
1
vote
1answer
97 views

Prove that in a triangle $\sum\limits_{cyc}\frac{w_bw_c}{w_a}\geq\frac{3}{4}\left(\sum\limits_{cyc}\frac{a^2w_a}{w_bw_c}\right)\geq\sqrt{3}s$

Let $ ABC$ is a triangle, $ w_a, w_b, w_c$ are bisectors of angles, $ h_a, h_b, h_c$ are altitudes respectively, $ r$ is radius of the incircle, prove that:$$ \frac {w_bw_c}{w_a} + \frac {w_cw_a}{w_b} ...
2
votes
1answer
104 views

An stronger inequality than in AoPS.

For $x,y,z >0.$ Prove$:$ $$\sum {\frac {y+z}{x}}+{\frac {1728 {x}^{ 3}{y}^{3}{z}^{3}}{ \left( x+y \right) ^{2} \left( y+z \right) ^{2} \left( z+x \right) ^{2} \left( x+y+z \right) ^{3}}} \geqslant ...
4
votes
1answer
110 views

An inequality involving real numbers

Let $x,y,z$ be real numbers such that $xyz=-1$. Prove that $$\sqrt[3/2]{\frac{3}{2}}\geq E:=\frac{4(x^3+y^3+z^3)}{(x^2+y^2+z^2)^2}$$ I tried to imitate an idea by River Li but it does not work. The ...
1
vote
1answer
64 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=...
2
votes
2answers
118 views

For which $k$ does $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+k-3\ge \left(2+\frac k3\right)\cdot \frac{a+b+c}{\sqrt[3]{abc}}$ hold?

By generalizing this (1) and this (2) questions and performing some research $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+k-3\ge \left(2+\frac k3\right)\cdot \frac{a+b+c}{\sqrt[3]{abc}},\...
0
votes
2answers
164 views

Given three real numbers $a,b,c$ so that $\{a, b, c\}\subset [1, 2]$ . Prove that $7abc\geq ab(a+ b)+ bc(b+ c)+ ca(c+ a)$ .

I need to a fresh solution with $a:\neq {\rm mid}\{a, b, c\}$ , but mine $$\begin{align*} 7abc &- ab(a+ b)- bc(b+ c)- ca(c+ a)= \\ &= a(2b- c)(2c- b)- (b+ c)(a- b)(a- c)\geq 0 \end{align*}$$...
2
votes
1answer
64 views

Let $a,b,c>0$ then prove this inequality holds

$$\left( \frac{a}{b+c}+\frac{b}{c+a} \right)\left( \frac{b}{c+a}+\frac{c}{a+b} \right)\left( \frac{c}{a+b}+\frac{a}{b+c} \right)\ge 1+\frac{1}{8}{{\left( \frac{a-b}{a+b} \right)}^{2}}{{\left( \frac{b-...
2
votes
1answer
89 views

Let $a,$ $b$ and $c$ are positive numbers.

Prove that $$\sqrt{\frac{a}{b+\alpha c}}+\sqrt{\frac{b}{c+\alpha a}}+\sqrt{\frac{c}{a+\alpha b}}\geq\frac{3}{\sqrt{1+\alpha}}$$ is true for all $\alpha\geq\dfrac{49+9\sqrt{17}}{32}$ I found this ...
3
votes
2answers
96 views

Let $a, b, c>0$. Prove that $\sum \limits_{cyc}{\frac{a}{b+c}\left(\frac{b}{c+a}+\frac{c}{a+b}\right)}\le \frac{(a+b+c)^2}{2(ab+bc+ca)}$

Reducing this whole expression i finally came to this $$\sum \limits_{cyc}\left(ab^4+a^4b+a^2b^2c\right)\geq \sum \limits_{cyc}\left(a^3b^2+a^2b^3+a^3bc\right)$$ Here I am stuck. I can't prove this. ...
2
votes
2answers
77 views

Prove$:$ $\sum\limits_{cyc} (\frac{a}{b+c}-\frac{1}{2}) \geqq (\sum\limits_{cyc} ab)\Big[\sum\limits_{cyc} \frac{1}{(a+b)^2}\Big]-\frac{9}{4}$

For $a,b,c$ are reals and $a+b+c>0, ab+bc+ca>0, (a+b)(b+c)(c+a)>0.$ Prove$:$ $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} -\frac{3}{2} \geqq (\sum\limits_{cyc} ab)\Big[\sum\limits_{cyc} \frac{...
0
votes
2answers
100 views

Proving inequality by SOS.

For $x,y,z>0.$ Prove$:$ $$P={x}^{4}y+{x}^{4}z+3\,{x}^{3}{y}^{2}-11\,{x}^{3}yz+3\,{x}^{3}{z}^{2}+3 \,{x}^{2}{y}^{3}+3\,{x}^{2}{y}^{2}z+3\,{x}^{2}y{z}^{2}+3\,{x}^{2}{z}^{ 3}+x{y}^{4}-11\,x{y}^{3}z+3\,...
6
votes
2answers
121 views

Prove $\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\geqq 1$

For $x,y,z>0$ and $\sqrt{x} +\sqrt{y} +\sqrt{z} =1.$ Prove that$:$ $$\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\geq 1$$ My solution$:$ Let $x=...
4
votes
2answers
122 views

Prove the following inequality $\sum_{i<j<k}\frac{a_ia_ja_k}{(n-2)(n-1)n}\le \bigg(\sum_{i<j}\frac{a_ia_j}{(n-1)n}\bigg)^2+\frac{1}{12}$

For positive integer $n \ge 3$, prove the following inequality $$\sum_{i<j<k}\frac{a_ia_ja_k}{(n-2)(n-1)n}\le \bigg(\sum_{i<j}\frac{a_ia_j}{(n-1)n}\bigg)^2+\frac{1}{12}$$ where $a_1+a_2+\...
2
votes
2answers
94 views

Proving $(a+b+c)^2\prod_{cyc}(a+b)-4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab\geqq 0$

From Mr. Michael Rozenberg solution: For $a,b,c>0$$,$ prove that$:$ $$(a+b+c)^2\prod_{cyc}(a+b)\geq4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab,$$ I found two SOS proof: 1) $$\text{LHS-RHS}={\frac { \left( ...
2
votes
2answers
111 views

Prove $\frac{a^2}{(b+c)^2}+\frac{b^2}{(c+a)^2}+\frac{c^2}{(a+b)^2}+\frac{1}{4}\ge \frac{a^2+b^2+c^2}{ab+bc+ca}$

For $a,b,c>0$. Prove that: $$\frac{a^2}{(b+c)^2}+\frac{b^2}{(c+a)^2}+\frac{c^2}{(a+b)^2}+\frac{1}{4}\geqq \frac{a^2+b^2+c^2}{ab+bc+ca}$$ NguyenHuyen gave the following expression$:$ $$\sum \...
3
votes
3answers
103 views

prove that $3(a+b+c) \geq 8(a b c)^{1 / 3}+\left(\frac{a^{3}+b^{3}+c^{3}}{3}\right)^{1 / 3}$

Question - Suppose a,b,c are positive real numbers , prove that $3(a+b+c) \geq 8(a b c)^{1 / 3}+\left(\frac{a^{3}+b^{3}+c^{3}}{3}\right)^{1 / 3}$ (Thailand $2006$) My attempt - we can assume ...
0
votes
1answer
90 views

Prove $\frac{3}{2} +\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \leqq \frac{a}{b}+\frac{b}{c} +\frac{c}{a}$

For $a,\,b,\,c>0$. Prove: $$\frac{3}{2} +\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \leqq \frac{a}{b}+\frac{b}{c} +\frac{c}{a}$$ My work: After a lot of caculates, I found: $\text{RHS-LHS}=$ ...
3
votes
3answers
108 views

show this inequality $\sum_{cyc}\frac{1}{5-2xy}\le 1$

let $x,y,z\ge 0$ and such $x^2+y^2+z^2=3$ show that $$\sum_{cyc}\dfrac{1}{5-2xy}\le 1$$ try: $$\sum_{cyc}\dfrac{2xy}{5-2xy}\le 2$$ and $$\sum_{cyc}\dfrac{2xy}{5-2xy}\le\sum_{cyc}\dfrac{(x+y)^2}{\frac{...
2
votes
3answers
107 views

Prove $\Big[\sum\limits_{cyc} a(a^2+2bc)\Big]^3 \geqq 3(ab+bc+ca)^2 . \sum\limits_{cyc} a(a^2+2bc)^2$

For $a,b,c>0$, prove that: $$ \Big[\sum\limits_{cyc} a(a^2+2bc)\Big]^3 \geqq 3(ab+bc+ca)^2 . \sum\limits_{cyc} a(a^2+2bc)^2$$ BW works here, but it's very ugly! My try: Let $p=a+b+c,q=ab+bc+ca,r=...
0
votes
4answers
127 views

MOP 2011 inequality

If $a,b,c$ are positive integers prove that $\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +\sqrt{(a^2+c^2-ac)} +9(abc)^{1/3} \le 4(a+b+c)$ My attempt: I tried to split inequality and prove it bit by ...

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