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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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Proving $5(a^2+b^2+c^2)+ab+bc+ca\ge 2\sum_{cyc}a\sqrt{5a^2+b^2+c^2+ab+ac}.$

For all $a,b,c\ge 0$ prove that$$\color{black}{5(a^2+b^2+c^2)+ab+bc+ca\ge 2\sum_{cyc}a\sqrt{5a^2+b^2+c^2+ab+ac}.}$$ I've tried to use AM-GM without success. Indeed, $$3\cdot RHS=2\sum_{cyc}3a\sqrt{5a^...
Sickness's user avatar
0 votes
2 answers
128 views

$a+b+c−3≥k(a−b)(b−c)(c−a)$

Given that $a,b,c \ge 0$ satisfy $ab+bc+ca=3$. Find the maximum value of the real number $k$ such that the following inequality is always true:$$a+b+c-3 \ge k(a-b)(b-c)(c-a)$$ "I got $c=0,ab=3 \...
POQ123's user avatar
  • 33
2 votes
2 answers
126 views

Proving $\sum_{cyc}\frac{a}{\sqrt{8a+bc}}\le \frac{\sqrt{a+b+c+abc}}{2}$ if $ab+bc+ca=3.$

Problem. Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=3.$ Prove that$$\color{black}{\frac{a}{\sqrt{8a+bc}}+\frac{b}{\sqrt{8b+ca}}+\frac{c}{\sqrt{8c+ab}}\le \frac{\sqrt{a+b+c+abc}}{2}.}$...
TATA box's user avatar
2 votes
1 answer
88 views

Proving $\sqrt{31a+b+c} +\sqrt{31b+a+c} +\sqrt{31c+b+a}\le 3\sqrt{3}\cdot\sqrt{a+b+c+8}$? when $a+b+c+abc=4.$

I came up with the inequality accidentally so there is no original proof so far. It would be great if you can give some useful help to prove it. Problem. Given non-negative real numbers $a,b,c$ ...
TATA box's user avatar
1 vote
2 answers
136 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
1 answer
86 views

Inequality sum of square root $3a^3+bc$ when $a^2+b^2+c^2=3$.

Remark. I really like inequality problems, but I've found that I have severe difficulties with ones that are very elegant form. I can prove something simple like this $$\color{black}{\sqrt{3a+bc}+\...
Sickness's user avatar
-2 votes
1 answer
98 views

How to solve $\sqrt{\frac{a}{bc+2}}+\sqrt{\frac{b}{ca+2}}+\sqrt{\frac{c}{ab+2}}\le \sqrt{\frac{3}{2}}\cdot\sqrt{a+b+c-1}$? [closed]

Let $a,b,c\ge 0: ab+bc+ca=3.$ Prove that $$\color{black}{\sqrt{\frac{a}{bc+2}}+\sqrt{\frac{b}{ca+2}}+\sqrt{\frac{c}{ab+2}}\le \sqrt{\frac{3}{2}}\cdot\sqrt{a+b+c-1}.}$$ Equality holds at $a=b=c=1.$ ...
Dragon boy's user avatar
0 votes
1 answer
166 views

Inequality $\frac{1}{\sqrt{ab+2}}+\frac{1}{\sqrt{bc+2}}+\frac{1}{\sqrt{ca+2}}\ge \frac{2\sqrt{3}}{\sqrt{a+b+c+abc}}.$

Let $a,b,c\ge 0: ab+bc+ca=3.$ Prove that $$\frac{1}{\sqrt{ab+2}}+\frac{1}{\sqrt{bc+2}}+\frac{1}{\sqrt{ca+2}}\ge \frac{2\sqrt{3}}{\sqrt{a+b+c+abc}}.$$ Equality occurs when $a=b=c=1.$ Also, there's an ...
Dragon boy's user avatar
2 votes
1 answer
74 views

Inequality $\frac{1}{\sqrt{4a^2+4b^2+17ab}}+\frac{1}{\sqrt{4b^2+4c^2+17bc}}+\frac{1}{\sqrt{4c^2+4a^2+17ca}}\ge \frac{3}{5}. $

Let $a,b,c\ge 0: ab+bc+ca+abc=4.$ Prove that $$\color{black}{\frac{1}{\sqrt{4a^2+4b^2+17ab}}+\frac{1}{\sqrt{4b^2+4c^2+17bc}}+\frac{1}{\sqrt{4c^2+4a^2+17ca}}\ge \frac{3}{5}. }$$ Equality holds at $a=b=...
Sickness's user avatar
0 votes
1 answer
83 views

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

For all $a,b,c\ge 0: ab+bc+ca>0$ then prove $$\color{black}{\sqrt{\frac{a}{b^2+bc+c^2}}+\sqrt{\frac{b}{c^2+ca+a^2}}+\sqrt{\frac{c}{a^2+ab+b^2}}\ge 2\sqrt{2}\sqrt{\frac{(ab+bc+ca)}{(a+b)(b+c)(c+a)}}....
Dragon boy's user avatar
0 votes
1 answer
72 views

Prove $\frac{bc}{\sqrt{7a+9}}+\frac{ca}{\sqrt{7b+9}}+\frac{ab}{\sqrt{7c+9}}\le \frac{3}{4}; a+b+c=3.$ [closed]

Given non-negative real numbers $a,b,c$ satisfying $a+b+c=3.$ Prove that $$\frac{bc}{\sqrt{7a+9}}+\frac{ca}{\sqrt{7b+9}}+\frac{ab}{\sqrt{7c+9}}\le \frac{3}{4}.$$ Equality holds at $a=b=c=1$ or $a=b=\...
Sickness's user avatar
0 votes
1 answer
57 views

Prove $\frac{bc}{\sqrt{3a+1}}+\frac{ca}{\sqrt{3b+1}}+\frac{ab}{\sqrt{3c+1}}\le \frac{3}{2}$ if $a^2+b^2+c^2=3.$

Given non-negative real numbers $a,b,c$ satisfying $a^2+b^2+c^2=3.$ Prove that $$\color{black}{\frac{bc}{\sqrt{3a+1}}+\frac{ca}{\sqrt{3b+1}}+\frac{ab}{\sqrt{3c+1}}\le \frac{3}{2}.}$$ Here is my ...
Dragon boy's user avatar
5 votes
3 answers
160 views

Prove $2(a+b+c)\left(1+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\ge 3(a+b)(b+c)(c+a)$ for $abc=1.$

Let $a,b,c>0: abc=1.$ Prove that: $$2(a+b+c)\left(1+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\ge 3(a+b)(b+c)(c+a). $$ I've tried to use a well-known lemma but the rest is quite complicated for me. ...
Anonymous's user avatar
1 vote
1 answer
112 views

Prove $\sqrt[3]{3a+bc}+\sqrt[3]{3b+ca}+\sqrt[3]{3c+ab}\le \frac{3}{2}\sqrt[3]{a^2+b^2+c^2+29}$ for $ab+bc+ca=3.$

If $a,b,c\ge 0: ab+bc+ca=3$ then prove $$\sqrt[3]{3a+bc}+\sqrt[3]{3b+ca}+\sqrt[3]{3c+ab}\le \frac{3}{2}\sqrt[3]{a^2+b^2+c^2+29}.$$ I tried to use AM-GM, Cauchy-Schwarz without success. Notice that $a^...
Dragon boy's user avatar
0 votes
2 answers
85 views

If $ab+bc+ca=3,$ prove $\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{c+b}}+\frac{1}{\sqrt{a+c}}\le \sqrt{\frac{2(a+b+c)+21}{6}}.$

If $a,b,c\ge 0: ab+bc+ca=3,$ then prove $$\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{c+b}}+\frac{1}{\sqrt{a+c}}\le \sqrt{\frac{2(a+b+c)+21}{6}}.$$ I've tried to use Cauchy-Schwarz and it's $$\frac{1}{a+b}+\...
Anonymous's user avatar
1 vote
3 answers
143 views

Prove $\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\ge \sqrt{2(ab+bc+ca)+12}$ when $a+b+c=3.$

Given non-negative real numbers $a,b,c$ satisfying $a+b+c=3.$ Prove that$$\color{black}{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\ge \sqrt{2(ab+bc+ca)+12}.}$$ Equality holds at $a=b=c=1$ or $a=b=0;c=3.$ I ...
Dragon boy's user avatar
8 votes
3 answers
271 views

How to prove $2\left(\sqrt{ab-1}+\sqrt{bc-1}+\sqrt{ca-1}\right)\le \left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\sqrt{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}.$?

Question. Let $a,b,c>0: abc=a+b+c+2.$ Prove that$$2\left(\sqrt{ab-1}+\sqrt{bc-1}+\sqrt{ca-1}\right)\le \left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\sqrt{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}.$$I am looking ...
Anonymous's user avatar
-1 votes
1 answer
91 views

How to prove $\sqrt{a+b}+\sqrt{c+b}+\sqrt{a+c}\le a+b+c+\frac{1}{2}abc+\sqrt{2}$ for $ab+bc+ca=1.$

Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=1.$ Prove that$$\color{black}{\sqrt{a+b}+\sqrt{c+b}+\sqrt{a+c}\le a+b+c+\frac{1}{2}abc+\sqrt{2}.}$$ Since $abc\ge 0$ and equality holds at ...
Sickness's user avatar
-1 votes
2 answers
129 views

Prove $\color{black}{\sqrt{\frac{7a+2}{b+c}}+\sqrt{\frac{7b+2}{a+c}}+\sqrt{\frac{7c+2}{b+a}}\ge \frac{9\sqrt{2}}{2} },$ if $a+b+c+abc=4.$

Problem. Let $a,b,c\ge 0: ab+bc+ca>0$ and $a+b+c+abc=4.$ Prove that $$\color{black}{\sqrt{\frac{7a+2}{b+c}}+\sqrt{\frac{7b+2}{a+c}}+\sqrt{\frac{7c+2}{b+a}}\ge \frac{9\sqrt{2}}{2} .}$$ Source: ...
Sickness's user avatar
0 votes
1 answer
73 views

Find minimal value $P=(a+b)(b+c)(c+a)$ if $ab+bc+ca=\sqrt{a}+\sqrt{b}+\sqrt{c}>0.$

Question. Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=\sqrt{a}+\sqrt{b}+\sqrt{c}>0.$ Find the minimal value of $P$ $$P=(a+b)(b+c)(c+a).$$ I check that Mininum is $8$ achieved at $...
Dragon boy's user avatar
1 vote
2 answers
90 views

Prove $\color{black}{\frac{1}{\sqrt{a+b+2}}+\frac{1}{\sqrt{c+b+2}}+\frac{1}{\sqrt{a+c+2}}\ge1+\frac{1}{\sqrt{2(a+b+c-1)}}.}$

Question. Let $a,b,c\ge 0: a+b+c+abc=4.$ Prove that$$\color{black}{\frac{1}{\sqrt{a+b+2}}+\frac{1}{\sqrt{c+b+2}}+\frac{1}{\sqrt{a+c+2}}\ge1+\frac{1}{\sqrt{2(a+b+c-1)}}.}$$ Because equality holds at $a=...
Dragon boy's user avatar
1 vote
2 answers
151 views

Given that $a,b,c>0$ and $abc=1$, prove that $a+b+c+\frac{3}{ab+bc+ca} \geq 4$

I was given some exercises from Math olympiads, and I am stuck with the one below, which seems soluble, yet I can't come up with something that works. Given that $a,b,c>0$ and $abc=1$, prove that $...
Limsup's user avatar
  • 396
2 votes
1 answer
75 views

$\sum \frac{a^2+b^2}{c^2}-\frac{7}{2} \geq \sqrt[3]{\prod \left(\frac{a}{b+c}+\frac{1}{3}\right)}$ for a,b,c real

Let: $a,b,c>0$. Prove that: $$\frac{a^2+b^2}{c^2}+\frac{b^2+c^2}{a^2}+\frac{c^2+a^2}{b^2}-\frac{7}{2}\geq \sqrt[3]{\left(\frac{a}{b+c}+\frac{1}{3}\right)\left(\frac{b}{c+a}+\frac{1}{3}\right)\left(\...
Lục Trường Phát's user avatar
0 votes
3 answers
112 views

Prove $\sqrt{\frac{a+b}{c+ab}}+\sqrt{\frac{c+b}{a+cb}}+\sqrt{\frac{a+c}{b+ac}}\ge 3,$ if $ab+bc+ca=3.$

If $a,b,c>0: ab+bc+ca=3,$ then prove that$$\sqrt{\frac{a+b}{c+ab}}+\sqrt{\frac{c+b}{a+cb}}+\sqrt{\frac{a+c}{b+ac}}\ge 3.$$ I've tried to use AM-GM without success. Indeed, we need to prove $$(a+b)(...
Dragon boy's user avatar
1 vote
1 answer
86 views

Prove that $\frac{1}{\sqrt{4a+bc}}+\frac{1}{\sqrt{4b+ca}}+\frac{1}{\sqrt{4c+ab}}\le \sqrt{2(a+b+c)}.$

If $a,b,c\ge 0: ab+bc+ca=1,$ then prove that $$\frac{1}{\sqrt{4a+bc}}+\frac{1}{\sqrt{4b+ca}}+\frac{1}{\sqrt{4c+ab}}\le \sqrt{2(a+b+c)}.$$ I saw it here. I think the following will help. Squaring both ...
Dragon boy's user avatar
5 votes
1 answer
132 views

Prove $\frac{a}{\sqrt{4a+3bc}}+\frac{b}{\sqrt{4b+3ca}}+\frac{c}{\sqrt{4c+3ab}}\le \frac{\sqrt{a+b+c+2}}{2}.$

Problem. Let $a,b,c\ge 0: ab+bc+ca=1.$ Prove that $$\frac{a}{\sqrt{4a+3bc}}+\frac{b}{\sqrt{4b+3ca}}+\frac{c}{\sqrt{4c+3ab}}\le \frac{\sqrt{a+b+c+2}}{2}.$$ Equality holds at $abc=0.$ I've tried to use ...
Anonymous's user avatar
3 votes
2 answers
144 views

$\frac34\le\sum_\text{cyc}\big(\frac{x}{x+1}\big)^2<2\ $ for $x,y,z>0$ and $xyz=1$.

For $x,y,z>0$ and $xyz=1$, what is the range of $f(x,y,z)$? $$f(x,y,z):=\Big(\frac{x}{x+1}\Big)^2+\Big(\frac{y}{y+1}\Big)^2+\Big(\frac{z}{z+1}\Big)^2.$$ I conjecture $\min f=\frac34$ at $(x,y,z)=(1,...
Hans's user avatar
  • 9,875
1 vote
1 answer
83 views

Simple proof to find minimum $P=(a^3+1)(b^3+1)(c^3+1)$

Let $a,b,c\ge 0: ab+bc+ca=abc+2.$ Find the minimum$$P=(a^3+1)(b^3+1)(c^3+1)$$By $a=b=c=1,$ we will prove $P\ge 8$ or $$(a^3+1)(b^3+1)(c^3+1)\ge 8.$$ After full expanding, it remains to prove $$(abc)^2+...
Dragon boy's user avatar
1 vote
2 answers
66 views

What is min $T=\sqrt{\frac{1-bc}{2b^2+5bc+2c^2}}+\sqrt{\frac{1-ac}{2a^2+5ac+2c^2}}+\sqrt{\frac{1-ba}{2b^2+5ba+2a^2}}$

Find the minimal value $$P=\sqrt{\frac{1-bc}{2b^2+5bc+2c^2}}+\sqrt{\frac{1-ac}{2a^2+5ac+2c^2}}+\sqrt{\frac{1-ba}{2b^2+5ba+2a^2}}$$ when $a,b,c\ge 0: ab+bc+ca=1.$ I set $a=b=1;c=0$ I got $P=\sqrt{2}.$ ...
Dragon boy's user avatar
1 vote
3 answers
134 views

How to prove $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge \frac{3}{2}$ with pqr?

Let $a,b,c\ge 0: a+b+c+3abc=6.$ Prove that $$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge \frac{3}{2}$$ I tried to expand $$\frac{2(a+b+c)+ab+bc+ca+3}{abc+ab+bc+ca+a+b+c+1}\ge \frac{3}{2}$$ $$\iff a+b+...
Anonymous's user avatar
0 votes
1 answer
43 views

Find max $T={\sum\frac {1-2yz}{3\,{y}^{2}+3\,{z}^{2}+{x}^{2}}}.$

Let $x,y,z\ge 0: xy+yz+zx=1.$ Find the maximum value $$T={\frac {1-2yz}{3\,{y}^{2}+3\,{z}^{2}+{x}^{2}}}+{\frac {1-2zx }{3\,{z}^{2}+3\,{x}^{2}+{y}^{2}}}+{\frac {1-2xy}{3\,{x}^{2}+3\,{y}^ {2}+{z}^{2}}}.$...
Anonymous's user avatar
2 votes
3 answers
87 views

Find max $P=a^3b^2+b^3c^2+c^3a^2+7(ab+bc+ca).$

Let $a,b,c\ge 0: a+b+c=3.$ Find the maximum of $$P=a^3b^2+b^3c^2+c^3a^2+7(ab+bc+ca)\le 24.$$ I set $a=b=c=1$ and got $P=24.$ My approach is proving $$a^3b^2+b^3c^2+c^3a^2+7(ab+bc+ca)\le 24$$ Firstly, ...
Anonymous's user avatar
1 vote
3 answers
109 views

If $x,y,z>0: x+y+z=3,$ prove $\frac{x^2}{y^2+yz+z^2}+\frac{y^2}{x^2+xz+z^2}+\frac{z^2}{y^2+yx+x^2}\ge \frac{5}{3}(x^2+y^2+z^2)+2xyz-6.$

Let $x,y,z>0: x+y+z=3.$ Prove that $$\frac{x^2}{y^2+yz+z^2}+\frac{y^2}{x^2+xz+z^2}+\frac{z^2}{y^2+yx+x^2}\ge \frac{5}{3}(x^2+y^2+z^2)+2xyz-6.$$ I tried to prove the well-known result $$\frac{x^2}{y^...
Anonymous's user avatar
1 vote
4 answers
115 views

For $x+y+z=3,$ prove $\frac{1}{x^2+4}+\frac{1}{y^2+4}+\frac{1}{z^2+4}\le \frac{3}{5}.$

Let $x,y,z\ge 0: x+y+z=3.$ Prove that$$\frac{1}{x^2+4}+\frac{1}{y^2+4}+\frac{1}{z^2+4}\le \frac{3}{5}.$$ Here is just my thought progress. I set $0\le xy+yz+zx=q\le 3; 0\le r=xyz\le 1.$ After full ...
Anonymous's user avatar
2 votes
2 answers
69 views

Finding $\small{\max\limits_{ab+bc+ca=abc+2}\frac{ab(2-c)}{a^2+abc+b^2}+\frac{bc(2-a)}{b^2+abc+c^2}+\frac{ca(2-b)}{c^2+abc+a^2}.}$

If $a,b,c\ge 0: ab+bc+ca=abc+2.$ Find the maximum $$P=\frac{ab(2-c)}{a^2+abc+b^2}+\frac{bc(2-a)}{b^2+abc+c^2}+\frac{ca(2-b)}{c^2+abc+a^2}$$ By set $a=b=c=1,$ we can see that $P= 1$. The remain is ...
Dragon boy's user avatar
2 votes
5 answers
469 views

Let $a+b+c=3,$ then prove$\sqrt[3]{\frac{a+b}{5ab+4}}+\sqrt[3]{\frac{c+b}{5cb+4}}+\sqrt[3]{\frac{a+c}{5ac+4}}\ge \sqrt[3]{6}$

Problem If $a,b,c\ge 0: a+b+c=3,$ then $$\sqrt[3]{\frac{a+b}{5ab+4}}+\sqrt[3]{\frac{c+b}{5cb+4}}+\sqrt[3]{\frac{a+c}{5ac+4}}\ge \sqrt[3]{6}.\tag{1}$$ I came up with when trying to prove $$\frac{a+b}{...
Sickness's user avatar
1 vote
3 answers
69 views

If $a^2+b^2+c^2=1$ prove $\frac{a^4-a^2}{bc-1}+\frac{b^4-b^2}{ca-1}+\frac{c^4-c^2}{ab-1}\le ab+bc+ca$

Let $a,b,c\ge 0: a^2+b^2+c^2=1.$ Prove that $$\frac{a^4-a^2}{bc-1}+\frac{b^4-b^2}{ca-1}+\frac{c^4-c^2}{ab-1}\le ab+bc+ca$$ Here is just my thought: After clear denominator, it suffices to prove $$\...
Dragon boy's user avatar
3 votes
2 answers
185 views

A difficult minimum [closed]

Positive real numbers a, b, and c satisfy that $(a+b)(b+c)(c+a)=10$, find the minimum value of the following $$(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2)+24abc$$ I have tried to square it completely $$ \...
Pesilk's user avatar
  • 121
2 votes
3 answers
122 views

Finding $\small{\min\limits_{ab+bc+ca=3}\dfrac{1}{4a+bc}+\dfrac{1}{4b+ca}+\dfrac{1}{4c+ab}.}$

Let $a,b,c> 0 :ab+bc+ca=3.$ Find the minimum$$P=\dfrac{1}{4a+bc}+\dfrac{1}{4b+ca}+\dfrac{1}{4c+ab}$$I am looking for a nice proof by hand, for which there is a possibility to find this proof during ...
Anonymous's user avatar
2 votes
4 answers
186 views

Finding $\small{\max\limits_{ab+bc+ca=3}\frac{ab}{2a+b}+\frac{bc}{2b+c}+\frac{ca}{2c+a}.}$

Let $a,b,c \ge 0: ab+bc+ca=3.$ Find maximal value $$P=\frac{ab}{2a+b}+\frac{bc}{2b+c}+\frac{ca}{2c+a}.$$ My tryings lead to wrong inequality. I tried to use C-S $$\sum_{cyc}\frac{ab}{2a+b}\le \sum_{...
Anonymous's user avatar
4 votes
4 answers
142 views

If $x^2+y^2+z^2=1,$ prove $\frac{x(z^2-y^2+2y)}{z}+\frac{y(x^2-z^2+2z)}{x}+\frac{z(y^2-x^2+2x)}{y}\ge 2$

Let $x,y,z>0: x^2+y^2+z^2=1.$ Prove that$$\frac{x(z^2-y^2+2y)}{z}+\frac{y(x^2-z^2+2z)}{x}+\frac{z(y^2-x^2+2x)}{y}\ge 2$$ I tried to use AM-GM $$\frac{x(z^2-y^2+2y)}{z}+\frac{y(x^2-z^2+2z)}{x}+\frac{...
Sickness's user avatar
1 vote
1 answer
42 views

$(a^2+b^2+c^2+k(ab+bc+ca))(\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2})\geq \frac{9}{4}(2-k)$

Let: $a,b,c \in \mathbb{R}$ and $a\neq b\neq c$. Prove that with $k \in [-1;2]$ we have: $(a^2+b^2+c^2+k(ab+bc+ca))(\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2})\geq \frac{9}{4}(2-k)$ By ...
Lục Trường Phát's user avatar
1 vote
2 answers
169 views

$ab+bc+ca+abc=4$, prove $\sum\frac{2+\sqrt{ab}}{\sqrt{ab}+c}+\frac{a^2+b^2+c^2}{8abc}\ge\frac{39}{8}$

Let $a,b,c> 0: ab+bc+ca+abc=4.$ Prove that: $$\frac{2+\sqrt{ab}}{\sqrt{ab}+c}+\frac{2+\sqrt{bc}}{\sqrt{bc}+a}+\frac{2+\sqrt{ca}}{\sqrt{ca}+b}+\frac{a^2+b^2+c^2}{8abc}\ge\frac{39}{8}.$$ I try a ...
Anonymous's user avatar
2 votes
3 answers
143 views

If $a+b+c=3,$ find max $T=\sqrt{\frac{6a+7bc}{6a+7}}+\sqrt{\frac{6b+7ca}{6b+7}}+\sqrt{\frac{6c+7ab}{6c+7}}$

Let $a,b,c\ge 0: a+b+c=3.$ Find the maximal value $$T=\sqrt{\frac{6a+7bc}{6a+7}}+\sqrt{\frac{6b+7ca}{6b+7}}+\sqrt{\frac{6c+7ab}{6c+7}}.$$ By $a=b=c=1,$ I get $M=3$ and I tried to prove it is maximum. ...
Dragon boy's user avatar
3 votes
3 answers
92 views

Prove $\sum\dfrac{b+c}{a+bc}\ge 2\left[(a+b)(b+c)(c+a)+\frac{3abc}{a+b+c}\right]$ for $ab+bc+ca=1$

Let $a,b,c\ge 0: ab+bc+ca=1.$ Prove$$\dfrac{b+c}{a+bc}+\dfrac{c+a}{b+ca}+\dfrac{a+b}{c+ab}\ge 2\left((a+b)(b+c)(c+a)+\frac{3abc}{a+b+c}\right) $$ I try C-S $$LHS\ge \frac{4(a+b+c)^2}{2+a+b+c-3abc}$$ ...
Anonymous's user avatar
1 vote
3 answers
119 views

How to prove $\frac{5abc+1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1\right)+6\ge \sum_{\text{cyc}}\sqrt{5a^3+4}$ when $a^2+b^2+c^2=a+b+c$

Question Let $a,b,c>0: a^2+b^2+c^2=a+b+c.$ Prove that $$\color{black}{\frac{5abc+1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1\right)+6\ge \sqrt{5a^3+4}+\sqrt{5b^3+4}+\sqrt{5c^3+4}}.$$ I've ...
Dragon boy's user avatar
0 votes
1 answer
79 views

Prove $abc+\frac{1}{ab+bc+ca}\ge 1$ when $a+b+c=2.$

If $a,b,c\ge 0:ab+bc+ca>0$ which sum is equal to $2$. Prove $$abc+\frac{1}{ab+bc+ca}\ge 1$$ I've tried to use Schur without success. I am waiting a proof without $uvw$ Let $0\le r\le \dfrac{8}{27}$ ...
Dragon boy's user avatar
0 votes
2 answers
91 views

Find the minimal constant $k$ :$\frac{a+k}{a+bc}+\frac{b+k}{b+ca}+\frac{c+k}{c+ab}\ge 3k+2,$ holds $ab+bc+ca=1.$

Problem. Find the minimal constant $k$ sastisfying $$\frac{a+k}{a+bc}+\frac{b+k}{b+ca}+\frac{c+k}{c+ab}\ge 3k+2,$$holds for all $a,b,c: ab+bc+ca=1.$ I've tried more but there is no good approach. ...
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2 votes
2 answers
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If $abc=25,$ find maximum $M=\frac{1}{b+c+12a}+\frac{1}{c+a+12b}+\frac{1}{a+b+12c}.$

Problem. Let $a,b,c>0:abc=25.$ Find maximum $$M=\frac{1}{b+c+12a}+\frac{1}{c+a+12b}+\frac{1}{a+b+12c}.$$ I've tried a new method $uvw$ When I set $a=b=5;c=1,$ calculate and I get $M=\dfrac{5}{66}....
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0 votes
2 answers
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Prove $2(xy+yz+zx)+3\ge \sqrt{5x^2+4}+\sqrt{5y^2+4}+\sqrt{5z^2+4}$ when $x,y,z>0: x+y+z=3xyz.$

Question Let $x,y,z>0: x+y+z=3xyz.$ Prove that$$2(xy+yz+zx)+3\ge \sqrt{5x^2+4}+\sqrt{5y^2+4}+\sqrt{5z^2+4}$$ I tried some classical inequality as AM-GM, Cauchy-Schwarz, etc but non of them work. ...
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