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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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 $$(abc)^2+...
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What is min $T=\sqrt{\frac{1-bc}{2b^2+5bc+2c^2}}+\sqrt{\frac{1-ac}{2a^2+5ac+2c^2}}+\sqrt{\frac{1-ba}{2b^2+5ba+2a^2}}$
Find the minimal value $$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}}$$
when $a,b,c\ge 0: ab+bc+ca=1.$
I set $a=b=1;c=0$ I got $P=\sqrt{2}.$
...
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How to prove $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge \frac{3}{2}$ with pqr?
Let $a,b,c\ge 0: a+b+c+3abc=6.$ Prove that $$\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+...
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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 $$T={\frac {1-2yz}{3\,{y}^{2}+3\,{z}^{2}+{x}^{2}}}+{\frac {1-2zx
}{3\,{z}^{2}+3\,{x}^{2}+{y}^{2}}}+{\frac {1-2xy}{3\,{x}^{2}+3\,{y}^
{2}+{z}^{2}}}.$...
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Find max $P=a^3b^2+b^3c^2+c^3a^2+7(ab+bc+ca).$
Let $a,b,c\ge 0: a+b+c=3.$ Find the maximum of $$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$$
Firstly, ...
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If $x,y,z>0: x+y+z=3,$ prove $\frac{x^2}{y^2+yz+z^2}+\frac{y^2}{x^2+xz+z^2}+\frac{z^2}{y^2+yx+x^2}\ge \frac{5}{3}(x^2+y^2+z^2)+2xyz-6.$
Let $x,y,z>0: x+y+z=3.$ Prove that $$\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^...
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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 ...
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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 ...
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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 $$\frac{a+b}{...
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If $a^2+b^2+c^2=1$ prove $\frac{a^4-a^2}{bc-1}+\frac{b^4-b^2}{ca-1}+\frac{c^4-c^2}{ab-1}\le ab+bc+ca$
Let $a,b,c\ge 0: a^2+b^2+c^2=1.$ Prove that $$\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$$
Here is just my thought:
After clear denominator, it suffices to prove $$\...
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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
$$
\...
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Finding $\small{\min\limits_{ab+bc+ca=3}\dfrac{1}{4a+bc}+\dfrac{1}{4b+ca}+\dfrac{1}{4c+ab}.}$
Let $a,b,c> 0 :ab+bc+ca=3.$ Find the minimum$$P=\dfrac{1}{4a+bc}+\dfrac{1}{4b+ca}+\dfrac{1}{4c+ab}$$I am looking for a nice proof by hand, for which there is a possibility to find this proof during ...
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Finding $\small{\max\limits_{ab+bc+ca=3}\frac{ab}{2a+b}+\frac{bc}{2b+c}+\frac{ca}{2c+a}.}$
Let $a,b,c \ge 0: ab+bc+ca=3.$ Find maximal value $$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_{...
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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 $$\frac{x(z^2-y^2+2y)}{z}+\frac{y(x^2-z^2+2z)}{x}+\frac{...
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$(a^2+b^2+c^2+k(ab+bc+ca))(\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2})\geq \frac{9}{4}(2-k)$
Let: $a,b,c \in \mathbb{R}$ and $a\neq b\neq c$. Prove that with $k \in [-1;2]$ we have:
$(a^2+b^2+c^2+k(ab+bc+ca))(\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2})\geq \frac{9}{4}(2-k)$
By ...
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$ab+bc+ca+abc=4$, prove $\sum\frac{2+\sqrt{ab}}{\sqrt{ab}+c}+\frac{a^2+b^2+c^2}{8abc}\ge\frac{39}{8}$
Let $a,b,c> 0: ab+bc+ca+abc=4.$ Prove that: $$\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}$$
I try a well-...
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If $a+b+c=3,$ find max $T=\sqrt{\frac{6a+7bc}{6a+7}}+\sqrt{\frac{6b+7ca}{6b+7}}+\sqrt{\frac{6c+7ab}{6c+7}}$
Let $a,b,c\ge 0: a+b+c=3.$ Find the maximal value $$T=\sqrt{\frac{6a+7bc}{6a+7}}+\sqrt{\frac{6b+7ca}{6b+7}}+\sqrt{\frac{6c+7ab}{6c+7}}.$$
By $a=b=c=1,$ I get $M=3$ and I tried to prove it is maximum.
...
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Prove $\sum\dfrac{b+c}{a+bc}\ge 2\left[(a+b)(b+c)(c+a)+\frac{3abc}{a+b+c}\right]$ for $ab+bc+ca=1$
Let $a,b,c\ge 0: ab+bc+ca=1.$ Prove$$\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}$$
...
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How to prove $\frac{5abc+1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1\right)+6\ge \sum_{\text{cyc}}\sqrt{5a^3+4}$ when $a^2+b^2+c^2=a+b+c$
Question
Let $a,b,c>0: a^2+b^2+c^2=a+b+c.$ Prove that $$\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}}.$$
I've ...
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Prove $abc+\frac{1}{ab+bc+ca}\ge 1$ when $a+b+c=2.$
If $a,b,c\ge 0:ab+bc+ca>0$ which sum is equal to $2$. Prove $$abc+\frac{1}{ab+bc+ca}\ge 1$$
I've tried to use Schur without success. I am waiting a proof without $uvw$
Let $0\le r\le \dfrac{8}{27}$ ...
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Find the minimal constant $k$ :$\frac{a+k}{a+bc}+\frac{b+k}{b+ca}+\frac{c+k}{c+ab}\ge 3k+2,$ holds $ab+bc+ca=1.$
Problem. Find the minimal constant $k$ sastisfying $$\frac{a+k}{a+bc}+\frac{b+k}{b+ca}+\frac{c+k}{c+ab}\ge 3k+2,$$holds for all $a,b,c: ab+bc+ca=1.$
I've tried more but there is no good approach. ...
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If $abc=25,$ find maximum $M=\frac{1}{b+c+12a}+\frac{1}{c+a+12b}+\frac{1}{a+b+12c}.$
Problem. Let $a,b,c>0:abc=25.$ Find maximum $$M=\frac{1}{b+c+12a}+\frac{1}{c+a+12b}+\frac{1}{a+b+12c}.$$
I've tried a new method $uvw$
When I set $a=b=5;c=1,$ calculate and I get $M=\dfrac{5}{66}....
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Prove $2(xy+yz+zx)+3\ge \sqrt{5x^2+4}+\sqrt{5y^2+4}+\sqrt{5z^2+4}$ when $x,y,z>0: x+y+z=3xyz.$
Question
Let $x,y,z>0: x+y+z=3xyz.$ Prove that$$2(xy+yz+zx)+3\ge \sqrt{5x^2+4}+\sqrt{5y^2+4}+\sqrt{5z^2+4}$$
I tried some classical inequality as AM-GM, Cauchy-Schwarz, etc but non of them work.
...
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If $ab+bc+ca=1,$ prove $1+36(abc)^2\ge\frac{21abc}{a+b+c}. $
Problem. Let $a,b,c\ge 0: ab+bc+ca=1.$ Prove that:$$1+36(abc)^2\ge\frac{21abc}{a+b+c}. $$
I think the problem is not hard but I am still stuck to find nice proofs.
After homogenizing, we need to ...
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If $ab+bc+ca=3$, then prove that $\sum\sqrt{\frac{b+c}{bc+1}}\ge\frac{a+b+c+3}{\sqrt{a+b+c+abc}}$
Problem Let $a,b,c\ge 0: ab+bc+ca=3$. Prove that$$\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}}.$$
I saw the problem on AOPS.
I tried to ...
- 1,047
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a+b+c=3. Prove that $\sqrt{(ab+2)(a+b)}+\sqrt{(bc+2)(b+c)}+\sqrt{(ca+2)(c+a)}\ge \frac{3\sqrt{3}}{2}\sqrt{3(ab+bc+ca)-abc}.$
Let $a,b,c\ge 0: a+b+c=3.$ Prove that$$\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}.$$
This problem is from a book.
I tried to used AM-GM ...
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Prove $\frac{1}{\sqrt{a+8b}}+\frac{1}{\sqrt{b+8c}}+\frac{1}{\sqrt{c+8a}}\ge 1$ for $ab + bc + ca = 3$
Let $a,b,c\ge 0: ab+bc+ca=3.$ Prove that$$\frac{1}{\sqrt{a+8b}}+\frac{1}{\sqrt{b+8c}}+\frac{1}{\sqrt{c+8a}}\ge 1.$$
This problem is from a book.
This cyclic inequality becomes an equality at $a=b=c=1.$...
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a,b,c>0 and prove $\sum\frac{\sqrt{a+b}}{c+\sqrt{ab}}\ge \frac{3\sqrt{2}}{2}\sqrt{\frac{a+b+c}{ab+bc+ca}}.$
Let $a,b,c>0$. Prove that: $$\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}}.$$
I used AM-GM: $\sqrt{...
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Prove: $\sum\limits_{cyc}\frac{1}{b^2+bc+c^2} \ge \frac{2}{ab+bc+ca}+\frac{a+b+c}{3(a^3+b^3+c^3)}.$ with $a,b,c \ge 0 : ab+bc+ca>0.$
Let $a,b,c \ge 0 : ab+bc+ca >0.$ Prove that $$\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)}.$$
The big problem here is ...
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$a,b,c\ge 0: a+b+c+abc=4$. Prove that: $\sum_{cyc}\sqrt{\frac{a+b}{c+ab}}\ge 3.$
Problem. Let $a,b,c\ge 0: a+b+c+abc=4$. Prove that: $$\sqrt{\frac{a+b}{c+ab}}+\sqrt{\frac{b+c}{a+bc}}+\sqrt{\frac{c+a}{b+ca}}\ge 3.$$
I think the problem is nice and very hard. Equality holds iff $a=...
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2
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If $a,b,c\ge 0: ab+bc+ca=1$, find Min $P=\frac{1}{\sqrt{2a+bc}}+\frac{1}{\sqrt{2b+ca}}+\frac{1}{\sqrt{2c+ab}}$
Problem. Let $a,b,c\ge 0: ab+bc+ca=1$. Find Minimum of $P$: $$P=\frac{1}{\sqrt{2a+bc}}+\frac{1}{\sqrt{2b+ca}}+\frac{1}{\sqrt{2c+ab}}$$ The source of the problem is unknown.
Here is my attempt:
We ...
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How to prove: $ab+bc+ca+\sqrt[3]{abc}+20\ge 2\sum_{cyc}\sqrt{a(4a+ab+bc+ca)}$ if $a+b+c=5$
Problem 1. Let $a,b,c\ge 0: a+b+c=5$. Prove that:
$$ab+bc+ca+\sqrt[3]{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(\...
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1
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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 $...
user1034536
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6
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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.
$$\iff\left(...
user1034536
2
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2
answers
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$3$-var inequality: $\frac{bc}{\sqrt{a}+3}+\frac{ca}{\sqrt{b}+3}+\frac{ab}{\sqrt{c}+3} \leq \frac{3}{4}$ for $a+b+c=3$.
Problem: Let $a,b,c$ be positive numbers satisfied $a+b+c=3$. Prove that $$\dfrac{bc}{\sqrt{a}+3}+\dfrac{ca}{\sqrt{b}+3}+\dfrac{ab}{\sqrt{c}+3} \leq \dfrac{3}{4}$$
I've tried U.C.T method but it doesn'...
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Prove $\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}\geq x^3+y^3+z^3$ where $x,y,z>0$ and $x+y+z=3$.
I just began learning the uvw method and tried solving this problem using it.
I would like to get my work reviewed and my mistakes pointed out.
Prove,whenever $x,y,z \geq 0$ and $x+y+z=3$ :$$\frac{...
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Smallest positive $k$ such that $\sum_{cyc}\left(\frac{(k-1)(z+x)}{y}-\frac{3}{2}\right)(x-y)^2\geq 0$
Let $x,y,z$ be positive real numbers. Find the smallest positive $k$ such that
$$\sum_{cyc}\left(\frac{(k-1)(z+x)}{y}-\frac{3}{2}\right)(x-y)^2\geq 0$$
holds true for $x,y,z$.
I proved it for $k=\frac{...
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2
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Prove that $\frac{a^4+2a+b(2a^2+b)}{a^2(b^2+2a)}+\frac{b^4+2b+c(2b^2+c)}{b^2(c^2+2b)}+\frac{c^4+2c+a(2c^2+a)}{c^2(a^2+2c)}\ge 6$
Problem: Let $a,b,c>0: \sqrt{a^4+8a}+\sqrt{b^4+8b}+\sqrt{c^4+8c}=3(a^2+b^2+c^2).$ Prove that: $$\frac{a^4+2a+b(2a^2+b)}{a^2(b^2+2a)}+\frac{b^4+2b+c(2b^2+c)}{b^2(c^2+2b)}+\frac{c^4+2c+a(2c^2+a)}{c^2(...
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5
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Prove that: $2(a+b+c)+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\ge\sqrt{5ab+4ac}+\sqrt{5bc+4ba}+\sqrt{5ca+4cb}$
Problem: For $a,b,c\ge0: ab+bc+ca>0.$ Prove that: $$2(a+b+c)+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\ge\sqrt{5ab+4ac}+\sqrt{5bc+4ba}+\sqrt{5ca+4cb}$$
Recently, i have seen a post on AoPS link
My approach: ...
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1
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Inequality without BW and Uvw, just AM GM
Problem: Let $a,b,c\ge0: ab+bc+ca=1.$ Prove that: $$(a+b+c)(3-\sqrt{ab}-\sqrt{bc}-\sqrt{ca})+2\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})\ge4$$
I guess equality holds: two of them equal $1$ and one equal ...
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2
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How do I prove that $\sum_{cyc}\left(\dfrac{1}{x^2-xy+y^2}\right)+15\ge6(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})$ given $x,y,z > 0$ and $x+y+z=3$?
I tried to apply AM-GM inequality:
$$ \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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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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4
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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:
$$ \dfrac{1}{\sqrt{2a^2+5ab+2b^2}}+\dfrac{1}{\sqrt{2b^2+5bc+2c^2}}+\dfrac{1}{\sqrt{2c^2+5ca+2a^2}} \geq\sqrt{\frac{3}...
- 193k
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4
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Let $a,b,c$ be non-negative real numbers .Prove that : $ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $
Let $a,b,c$ be non-negative real numbers
Prove that : $ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $
My idea is to use the $(p,q,r)$ method:
$p=a+b+c$
$q=ab+bc+ca$
$r = abc $
$...
- 1,170
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5
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Find the maximum of the value $F=x^3y+y^3z+z^3x$
let $x,y,z$ be real number.if $x+y+z=3$,show that
$$x^3y+y^3z+z^3x\le \dfrac{9(63+5\sqrt{105})}{32}$$
and the inequality $=$,then $x=?,y=?,z=?$
I can solve if add $x,y,z\ge 0$,also see: Calculate ...
- 92.5k
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Hard and non-trivial inequality with three variable reals
For all reals $a$, $b$, $c$, show that $$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]{\...
- 131
3
votes
1
answer
185
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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 $...
- 349
7
votes
1
answer
161
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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:
$$3\sqrt[3]{a}+3\sqrt[3]{b}+3\sqrt[3]{c}+15\ge 3(a+b)(b+...
- 153
0
votes
0
answers
56
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With the $uwv$ method, why can $3v^2$ be negative?
If we define $3v^2 = ab+bc+ca$, then doesn't that mean $3v^2$ must be positive due to the exponentiation and positive $3$?
I note that if for example $a = −1, b = −2, c = 3$ we get $3v^2 = −7$! I ...
- 1,533
5
votes
3
answers
116
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Prove that: $(a^5-2a+4)(b^5-2b+4)(c^5-2c+4)\ge9(ab+bc+ca)$
Let $a,b,c>0$ satisfy $a^2+b^2+c^2=3$ . Prove that: $$(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(...
- 81