# 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.

254 questions
Filter by
Sorted by
Tagged with
71 views

36 views

95 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 ...
66 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 ...
137 views

178 views

119 views

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 $$\frac{x(z^2-y^2+2y)}{z}+\frac{y(x^2-z^2+2z)}{x}+\frac{... 1 vote 1 answer 37 views ### (a^2+b^2+c^2+k(ab+bc+ca))(\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2})\geq \frac{9}{4}(2-k) Let: a,b,c \in \mathbb{R} and a\neq b\neq c. Prove that with k \in [-1;2] we have: (a^2+b^2+c^2+k(ab+bc+ca))(\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2})\geq \frac{9}{4}(2-k) By ... 0 votes 0 answers 38 views ### ab+bc+ca+abc=4, prove \sum\frac{2+\sqrt{ab}}{\sqrt{ab}+c}+\frac{a^2+b^2+c^2}{8abc}\ge\frac{39}{8} Let a,b,c> 0: ab+bc+ca+abc=4. Prove that:$$\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}$$I try a well-... 2 votes 1 answer 96 views ### If a+b+c=3, find max T=\sqrt{\frac{6a+7bc}{6a+7}}+\sqrt{\frac{6b+7ca}{6b+7}}+\sqrt{\frac{6c+7ab}{6c+7}} Let a,b,c\ge 0: a+b+c=3. Find the maximal value$$T=\sqrt{\frac{6a+7bc}{6a+7}}+\sqrt{\frac{6b+7ca}{6b+7}}+\sqrt{\frac{6c+7ab}{6c+7}}.$$By a=b=c=1, I get M=3 and I tried to prove it is maximum. ... 4 votes 3 answers 80 views ### Prove \sum\dfrac{b+c}{a+bc}\ge 2\left[(a+b)(b+c)(c+a)+\frac{3abc}{a+b+c}\right] for ab+bc+ca=1 Let a,b,c\ge 0: ab+bc+ca=1. Prove$$\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}$$... 2 votes 3 answers 106 views ### How to prove \frac{5abc+1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1\right)+6\ge \sum_{\text{cyc}}\sqrt{5a^3+4} when a^2+b^2+c^2=a+b+c Question Let a,b,c>0: a^2+b^2+c^2=a+b+c. Prove that$$\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}}.$$I've ... 0 votes 1 answer 69 views ### Prove abc+\frac{1}{ab+bc+ca}\ge 1 when a+b+c=2. If a,b,c\ge 0:ab+bc+ca>0 which sum is equal to 2. Prove$$abc+\frac{1}{ab+bc+ca}\ge 1$$I've tried to use Schur without success. I am waiting a proof without uvw Let 0\le r\le \dfrac{8}{27} ... 0 votes 2 answers 88 views ### Find the minimal constant k :\frac{a+k}{a+bc}+\frac{b+k}{b+ca}+\frac{c+k}{c+ab}\ge 3k+2, holds ab+bc+ca=1. Problem. Find the minimal constant k sastisfying$$\frac{a+k}{a+bc}+\frac{b+k}{b+ca}+\frac{c+k}{c+ab}\ge 3k+2,$$holds for all a,b,c: ab+bc+ca=1. I've tried more but there is no good approach. ... 2 votes 2 answers 179 views ### If abc=25, find maximum M=\frac{1}{b+c+12a}+\frac{1}{c+a+12b}+\frac{1}{a+b+12c}. Problem. Let a,b,c>0:abc=25. Find maximum$$M=\frac{1}{b+c+12a}+\frac{1}{c+a+12b}+\frac{1}{a+b+12c}.$$I've tried a new method uvw When I set a=b=5;c=1, calculate and I get M=\dfrac{5}{66}.... 0 votes 2 answers 61 views ### Prove 2(xy+yz+zx)+3\ge \sqrt{5x^2+4}+\sqrt{5y^2+4}+\sqrt{5z^2+4} when x,y,z>0: x+y+z=3xyz. Question Let x,y,z>0: x+y+z=3xyz. Prove that$$2(xy+yz+zx)+3\ge \sqrt{5x^2+4}+\sqrt{5y^2+4}+\sqrt{5z^2+4}$$I tried some classical inequality as AM-GM, Cauchy-Schwarz, etc but non of them work. ... 1 vote 3 answers 98 views ### If ab+bc+ca=1, prove 1+36(abc)^2\ge\frac{21abc}{a+b+c}.  Problem. Let a,b,c\ge 0: ab+bc+ca=1. Prove that:$$1+36(abc)^2\ge\frac{21abc}{a+b+c}. $$I think the problem is not hard but I am still stuck to find nice proofs. After homogenizing, we need to ... 1 vote 3 answers 123 views ### If ab+bc+ca=3, then prove that \sum\sqrt{\frac{b+c}{bc+1}}\ge\frac{a+b+c+3}{\sqrt{a+b+c+abc}} Problem Let a,b,c\ge 0: ab+bc+ca=3. Prove that$$\sqrt{\frac{b+c}{bc+1}}+\sqrt{\frac{c+a}{ca+1}}+\sqrt{\frac{a+b}{ab+1}}\ge\frac{a+b+c+3}{\sqrt{a+b+c+abc}}.$$I saw the problem on AOPS. I tried to ... 1 vote 1 answer 74 views ### a+b+c=3. Prove that \sqrt{(ab+2)(a+b)}+\sqrt{(bc+2)(b+c)}+\sqrt{(ca+2)(c+a)}\ge \frac{3\sqrt{3}}{2}\sqrt{3(ab+bc+ca)-abc}. Let a,b,c\ge 0: a+b+c=3. Prove that$$\sqrt{(ab+2)(a+b)}+\sqrt{(bc+2)(b+c)}+\sqrt{(ca+2)(c+a)}\ge \frac{3\sqrt{3}}{2}\sqrt{3(ab+bc+ca)-abc}.$$This problem is from a book. I tried to used AM-GM ... 3 votes 2 answers 179 views ### Prove \frac{1}{\sqrt{a+8b}}+\frac{1}{\sqrt{b+8c}}+\frac{1}{\sqrt{c+8a}}\ge 1 for ab + bc + ca = 3 Let a,b,c\ge 0: ab+bc+ca=3. Prove that$$\frac{1}{\sqrt{a+8b}}+\frac{1}{\sqrt{b+8c}}+\frac{1}{\sqrt{c+8a}}\ge 1.$$This problem is from a book. This cyclic inequality becomes an equality at a=b=c=1.... 0 votes 1 answer 69 views ### a,b,c>0 and prove \sum\frac{\sqrt{a+b}}{c+\sqrt{ab}}\ge \frac{3\sqrt{2}}{2}\sqrt{\frac{a+b+c}{ab+bc+ca}}. Let a,b,c>0. Prove that:$$\frac{\sqrt{a+b}}{c+\sqrt{ab}}+\frac{\sqrt{b+c}}{a+\sqrt{bc}}+\frac{\sqrt{c+a}}{b+\sqrt{ca}}\ge \frac{3\sqrt{2}}{2}\sqrt{\frac{a+b+c}{ab+bc+ca}}.$$I used AM-GM: \sqrt{... 1 vote 1 answer 66 views ### Prove: \sum\limits_{cyc}\frac{1}{b^2+bc+c^2} \ge \frac{2}{ab+bc+ca}+\frac{a+b+c}{3(a^3+b^3+c^3)}. with a,b,c \ge 0 : ab+bc+ca>0. Let a,b,c \ge 0 : ab+bc+ca >0. Prove that$$\dfrac{1}{a^2+ab+b^2}+\dfrac{1}{b^2+bc+c^2}+\dfrac{1}{c^2+ca+a^2} \ge \dfrac{2}{ab+bc+ca}+\dfrac{a+b+c}{3(a^3+b^3+c^3)}.$$The big problem here is ... 6 votes 4 answers 407 views ### a,b,c\ge 0: a+b+c+abc=4. Prove that: \sum_{cyc}\sqrt{\frac{a+b}{c+ab}}\ge 3. Problem. Let a,b,c\ge 0: a+b+c+abc=4. Prove that:$$\sqrt{\frac{a+b}{c+ab}}+\sqrt{\frac{b+c}{a+bc}}+\sqrt{\frac{c+a}{b+ca}}\ge 3.$$I think the problem is nice and very hard. Equality holds iff a=... 8 votes 2 answers 408 views ### If a,b,c\ge 0: ab+bc+ca=1, find Min P=\frac{1}{\sqrt{2a+bc}}+\frac{1}{\sqrt{2b+ca}}+\frac{1}{\sqrt{2c+ab}} Problem. Let a,b,c\ge 0: ab+bc+ca=1. Find Minimum of P:$$P=\frac{1}{\sqrt{2a+bc}}+\frac{1}{\sqrt{2b+ca}}+\frac{1}{\sqrt{2c+ab}}$$The source of the problem is unknown. Here is my attempt: We ... 5 votes 3 answers 2k views ### How to prove: ab+bc+ca+\sqrt{abc}+20\ge 2\sum_{cyc}\sqrt{a(4a+ab+bc+ca)} if a+b+c=5 Problem 1. Let a,b,c\ge 0: a+b+c=5. Prove that:$$ab+bc+ca+\sqrt{abc}+20 \ge 2\left(\sqrt{a(4a+ab+bc+ca)}+\sqrt{b(4b+ab+bc+ca)}+\sqrt{c(4c+ab+bc+ca)}\right)$$Equality holds iff (a,b,c)=\left(\... 2 votes 1 answer 122 views ### Maximize \sum a_i^3 for positive a_i's with \sum a_i=\sum a_i^2=n-1 If n positive real numbers a_i~(1\le i\le n) satisfy \sum\limits_{i=1}^na_i=\sum\limits_{i=1}^na_i^2=n-1, find the maximum or supremum of $M=a_1^3+a_2^3+\cdots+a_n^3.$ I think the maximum of ... 5 votes 6 answers 220 views ### Prove that (\sum x^2)^3\ge9\sum x^4yz Prove that \displaystyle\left(x^2+y^2+z^2\right)^3\ge9\left(x^4yz+y^4xz+z^4xy\right), for x, y, z\in\Bbb R_+. The pqr method doesn't seem possible because the power is too high.$$\iff\left(... 145 views

### $3$-var inequality: $\frac{bc}{\sqrt{a}+3}+\frac{ca}{\sqrt{b}+3}+\frac{ab}{\sqrt{c}+3} \leq \frac{3}{4}$ for $a+b+c=3$.

Problem: Let $a,b,c$ be positive numbers satisfied $a+b+c=3$. Prove that $$\dfrac{bc}{\sqrt{a}+3}+\dfrac{ca}{\sqrt{b}+3}+\dfrac{ab}{\sqrt{c}+3} \leq \dfrac{3}{4}$$ I've tried U.C.T method but it doesn'...
1 vote
72 views

220 views

### 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
116 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 ...