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

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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### 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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### 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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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 : $$\sum a^4b+3\... 6 votes 2 answers 132 views ### Find the inequality with the best possible k= constant (with the condition x^{2}+ y^{2}\leq k). Find the inequality with the best possible constant Given two non-negative numbers x, y so that x^{2}+ y^{2}\leq \frac{2}{7}. Prove that$$\frac{1}{1+ x^{2}}+ \frac{1}{1+ y^{2}}+ \frac{1}{1+ xy}... 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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### 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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By generalizing this (1) and this (2) questions and performing some research $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+k-3\ge \left(2+\frac k3\right)\cdot \frac{a+b+c}{\sqrt{abc}},\... 0 votes 2 answers 168 views ### Given three real numbers a,b,c so that \{a, b, c\}\subset [1, 2] . Prove that 7abc\geq ab(a+ b)+ bc(b+ c)+ ca(c+ a) . I need to a fresh solution with a:\neq {\rm mid}\{a, b, c\} , but mine$$\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*}... 