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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.

1 vote
2 answers
201 views

Let $a,b,c$ be real variables with $ab+bc+ca+abc=4.$ Prove that:$$\color{black}{\frac{7-4a}{a^{2}+2}+\frac{7-4b}{b^{2}+2}+\frac{7-4c}{c^{2}+2}\ge 3.}$$When does equality hold? This inequality is ...
Danh Trung's user avatar
0 votes
0 answers
123 views

Let $x, y, z$ be non-negative real numbers such that $x + y + z = 3$. Find the maximum value of $$ \frac{1}{x^2 + 1} + \frac{1}{y^2 + 1} + \frac{1}{z^2 + 1}. $$ My attempt: First, I assumed $z = 0$, ...
Student's user avatar
  • 11
-1 votes
1 answer
88 views

I'm looking for some ideas to solve the following inequality. Problem. Let $a,b,c\ge 0$ with $ab+bc+ca=1.$ Prove that$$\color{black}{\frac{a\left(b+c+2\right)}{bc+2a}+\frac{b\left(c+a+2\right)}{ca+2b}...
Danh Trung's user avatar
0 votes
1 answer
240 views

I am very interested in a motivated systematic approach to generating lemmas such as those invoked by River Li to transform inequalities stated in terms of radicals into inequalities that are radical ...
Phil's user avatar
  • 57
1 vote
3 answers
212 views

Problem. Let $a,b,c$ be positive real numbers such that $$(a+b+c)^3=125abc$$ Prove that : $$ \dfrac{a}{\sqrt{bc}}+\dfrac{b}{\sqrt{ca}}+\dfrac{c}{\sqrt{ab}} \le \dfrac{16+\sqrt{2}}{2}$$ This problem ...
Kazumi's user avatar
  • 111
0 votes
1 answer
302 views

I found a interesting problem in Aops. It is a hard inequality problem. Here is the problem and my approach(and the link): https://artofproblemsolving.com/community/c6h3188980p34729637 Let $a,b,c$ be ...
Batuhan Yılmaz's user avatar
0 votes
1 answer
142 views

I'm looking for some ideas to solve the following inequality. Problem. Let $a,b,c>0: a+b+c=3$ then prove that$$\frac{1}{\left(1+\sqrt{\dfrac{a}{bc}}\right)^2}+\frac{1}{\left(1+\sqrt{\dfrac{b}{ca}}\...
30 Anh Ti 711's user avatar
0 votes
2 answers
186 views

Prove $\dfrac{a}{\sqrt{b^2 + 3c^2}} + \dfrac{b}{\sqrt{c^2 + 3a^2}} + \dfrac{c}{\sqrt{a^2 + 3b^2}} \geq \dfrac{3}{2}$ if $a,b,c > 0$. I have tried using Cauchy-Schwarz and Holder, but nothing works ...
Đạt Dư Đình's user avatar
2 votes
1 answer
80 views

I'm looking for some ideas to solve the following inequality. Problem. For all non-negative real variables $a,b,c,$ prove that$$m^2+3pn\ge 5n^2+pm$$holds where$$p=a+b+c;\quad m=\sqrt{ab}+\sqrt{bc}+\...
30 Anh Ti 711's user avatar
2 votes
1 answer
89 views

I'm looking for some ideas to solve the following inequality. Problem. Let $a,b,c\ge 0: ab+bc+ca+abc=4$ then prove $$\color{black}{\frac{1}{4ab+4ac+3}+\frac{1}{4bc+4ba+3}+\frac{1}{4ca+4cb+3}\ge \...
30 Anh Ti 711's user avatar
3 votes
3 answers
377 views

I'm looking for some ideas to solve the following inequality. Problem. For any real numbers $a,b,c$ with $a+b+c=3,$ prove that$$\color{black}{\frac{a}{6a^{2}+a+2}+\frac{b}{6b^{2}+b+2}+\frac{c}{6c^{2}...
30 Anh Ti 711's user avatar
2 votes
2 answers
291 views

Let $a,$ $b$ and $c$ be non-negative numbers such that: $a+b+c+abc=4$ and $ab+bc+ca\neq 0.$ Prove that $$\frac{1}{\sqrt{a^2+4bc}}+\frac{1}{\sqrt{b^2+4ca}}+\frac{1}{\sqrt{c^2+4ab}}\ge \frac{5}{4}.$$ ...
Michael Rozenberg's user avatar
6 votes
1 answer
550 views

How do we prove that for all $a, b, c \in \mathbb{R}$, $$\frac{2a}{a^2+2b^2+3}+\frac{2b}{b^2+2c^2+3}+\frac{2c}{c^2+2a^2+3} \leq 1.$$ I haven't really made much progress in finding a way to tackle this....
MilesB's user avatar
  • 930
-1 votes
1 answer
139 views

I'm looking for some ideas to solve the following inequality. Problem. For any non-negative real numbers $a,b,c$ with $ab+bc+ca+abc=4,$ prove that$$\color{blue}{\sqrt{\frac{a}{bc+8}}+\sqrt{\frac{b}{...
Daniel Lest's user avatar
3 votes
0 answers
234 views

Let $a$, $b$ and $c$ be real numbers such that $a+b+c=3$. Prove that: $$\frac{ab+3c}{5c^2+4}+\frac{bc+3a}{5a^2+4}+\frac{ca+3b}{5b^2+4}\leq\frac{4}{3}.$$ This inequality was posted here. I solved it ...
Michael Rozenberg's user avatar

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