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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Prove that: $2(a+b+c)+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\ge\sqrt{5ab+4ac}+\sqrt{5bc+4ba}+\sqrt{5ca+4cb}$

Problem: For $a,b,c\ge0: ab+bc+ca>0.$ Prove that: $$2(a+b+c)+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\ge\sqrt{5ab+4ac}+\sqrt{5bc+4ba}+\sqrt{5ca+4cb}$$ Recently, i have seen a post on AoPS link My approach: ...
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1 vote
1 answer
62 views

Prove that: $\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}+\sqrt{3(a^2+b^2+c^2-ab-bc-ca)}\ge \frac{3}{2}$

Problem: Given non- negative real numbers such that: $ab+bc+ca>0: a+b+c=3.$ Prove that: $$\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}+\sqrt{3(a^2+b^2+c^2-ab-bc-ca)}\ge \frac{3}{2}$$ My ...
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  • 409
1 vote
2 answers
90 views

Inequality without BW and Uvw, just AM GM

Problem: Let $a,b,c\ge0: ab+bc+ca=1.$ Prove that: $$(a+b+c)(3-\sqrt{ab}-\sqrt{bc}-\sqrt{ca})+2\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})\ge4$$ I guess equality holds: two of them equal $1$ and one equal ...
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  • 409
2 votes
2 answers
100 views

How do I prove that $\sum_{cyc}\left(\dfrac{1}{x^2-xy+y^2}\right)+15\ge6(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})$ given $x,y,z > 0$ and $x+y+z=3$?

I tried to apply AM-GM inequality: $$ \dfrac{1}{x^2-xy+y^2} + (x^2-xy+y^2) \ge 2 \implies \dfrac{1}{x^2-xy+y^2} \ge 2 - (x^2-xy+y^2) $$ Then, $$ \sum_{cyc}\left(\dfrac{1}{x^2-xy+y^2}\right)+15\ge 21-2(...
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  • 51
2 votes
3 answers
156 views

Prove that $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} \ge ab + bc + ca$

For all positive $a,b,c $ satisfying $a+b+c = 3$,Prove: $$ \sum_{cyc} \sqrt[3]{a} \ge \sum_{cyc} ab $$ This is a hard problem and I tried it myself, but it's really hard without using advanced ...
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5 votes
3 answers
236 views

Prove that: $\sum\limits_{cyc}\frac{1}{\sqrt{2a^2+5ab+2b^2}} \geq\sqrt{\frac{3}{ab+ac+bc}}$

Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that: $$ \dfrac{1}{\sqrt{2a^2+5ab+2b^2}}+\dfrac{1}{\sqrt{2b^2+5bc+2c^2}}+\dfrac{1}{\sqrt{2c^2+5ca+2a^2}} \geq\sqrt{\frac{3}...
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3 votes
4 answers
150 views

Let $a,b,c$ be non-negative real numbers .Prove that : $ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $

Let $a,b,c$ be non-negative real numbers Prove that : $ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $ My idea is to use the $(p,q,r)$ method: $p=a+b+c$ $q=ab+bc+ca$ $r = abc $ $...
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  • 1,130
1 vote
5 answers
344 views

Find the maximum of the value $F=x^3y+y^3z+z^3x$

let $x,y,z$ be real number.if $x+y+z=3$,show that $$x^3y+y^3z+z^3x\le \dfrac{9(63+5\sqrt{105})}{32}$$ and the inequality $=$,then $x=?,y=?,z=?$ I can solve if add $x,y,z\ge 0$,also see: Calculate ...
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4 votes
2 answers
99 views

Hard and non-trivial inequality with three variable reals

For all reals $a$, $b$, $c$, show that $$a^2+b^2+c^2 \geq a\sqrt[\leftroot{-1}\uproot{1}4]{\frac{b^4+c^4}{2}} + b\sqrt[\leftroot{-1}\uproot{1}4]{\frac{c^4+a^4}{2}} + c\sqrt[\leftroot{-1}\uproot{1}4]{\...
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3 votes
1 answer
179 views

An inequality with $x,y,z \ge -1$ and $x+y+z=1$

For $x,y,z \ge -1$ and $x+y+z=1$, Prove: $$ \frac{8x}{yz}+\frac{8y}{zx}+\frac{8z}{xy}+\frac{1}{x^2y^2z^2}+9 \ge \frac{30}{xyz} $$ The difficulties of this qustion are that the constraint condition is $...
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7 votes
1 answer
160 views

Prove that: $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}+5\ge (a+b)(b+c)(c+a)$

Let $a,b,c>0$ satisfy $a+b+c=3$ Prove that: $$\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}+5\ge (a+b)(b+c)(c+a)$$I have my solution, we need to prove: $$3\sqrt[3]{a}+3\sqrt[3]{b}+3\sqrt[3]{c}+15\ge 3(a+b)(b+...
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0 votes
0 answers
41 views

With the $uwv$ method, why can $3v^2$ be negative?

If we define $3v^2 = ab+bc+ca$, then doesn't that mean $3v^2$ must be positive due to the exponentiation and positive $3$? I note that if for example $a = −1, b = −2, c = 3$ we get $3v^2 = −7$! I ...
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  • 1,495
5 votes
3 answers
105 views

Prove that: $(a^5-2a+4)(b^5-2b+4)(c^5-2c+4)\ge9(ab+bc+ca)$

Let $a,b,c>0$ satisfy $a^2+b^2+c^2=3$ . Prove that: $$(a^5-2a+4)(b^5-2b+4)(c^5-2c+4)\ge9(ab+bc+ca)$$ My idea is to use a well-known inequality (We can prove by Schur) $$(a^2+2)(b^2+2)(c^2+2)\ge 9(...
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  • 81
-4 votes
1 answer
150 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^...
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  • 3,515
14 votes
2 answers
544 views

The inequality $\,2+\sqrt{\frac p2}\leq\sum\limits_\text{cyc}\sqrt{\frac{a^2+pbc}{b^2+c^2}}\,$ where $0\leq p\leq 2$ is: Probably true! Provably true?

Let $p$ be a positive parameter in the range from $0$ to $2$. Can one prove that $$2 +\sqrt{\frac p2} \;\leqslant\;\sqrt{\frac{a^2 + pbc}{b^2+c^2}} \,+\,\sqrt{\frac{b^2 +pca}{c^2+a^2}}\,+\,\sqrt{\...
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  • 4,930
1 vote
0 answers
64 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 (...
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1 vote
1 answer
81 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.
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  • 1,950
10 votes
0 answers
303 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 ...
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  • 8,418
0 votes
0 answers
201 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}\...
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  • 3,515
2 votes
2 answers
75 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 $$P\le \sqrt{\frac{16}{27}}$$ $...
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1 vote
1 answer
108 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[(\...
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4 votes
3 answers
164 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)\...
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8 votes
3 answers
704 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 $a,b,c\geqslant0$ $$ \...
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2 votes
4 answers
158 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 \...
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  • 675
4 votes
1 answer
131 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 ...
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  • 909
0 votes
4 answers
130 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+...
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0 votes
1 answer
81 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}...
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  • 1,950
4 votes
3 answers
121 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{...
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  • 1,950
2 votes
2 answers
161 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^...
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  • 1,950
4 votes
2 answers
170 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 ...
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  • 1,950
4 votes
2 answers
135 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( \...
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  • 1,950
2 votes
3 answers
103 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 $a+b+c\le a^2+b^2+c^2$ ...
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2 votes
2 answers
104 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}}{...
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  • 1,950
3 votes
1 answer
154 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\...
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  • 1,950
0 votes
4 answers
145 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 ...
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3 votes
5 answers
126 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\...
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6 votes
2 answers
132 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}...
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4 votes
0 answers
153 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 ...
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  • 1,950
2 votes
4 answers
113 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}(...
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1 vote
1 answer
101 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 ...
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  • 1,950
1 vote
0 answers
100 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} \...
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  • 1,950
0 votes
3 answers
95 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 ...
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2 votes
7 answers
208 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({...
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2 votes
1 answer
152 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 ...
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  • 1,950
1 vote
1 answer
102 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} ...
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2 votes
1 answer
118 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 ...
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  • 1,950
5 votes
1 answer
126 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 ...
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  • 172
1 vote
1 answer
67 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=...
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  • 1,950
2 votes
2 answers
127 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}},\...
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0 votes
2 answers
168 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 $$\begin{align*} 7abc &- ab(a+ b)- bc(b+ c)- ca(c+ a)= \\ &= a(2b- c)(2c- b)- (b+ c)(a- b)(a- c)\geq 0 \end{align*}$$...
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