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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54 views

Proving inequality by SOS.

For $x,y,z>0.$ Prove$:$ $$P={x}^{4}y+{x}^{4}z+3\,{x}^{3}{y}^{2}-11\,{x}^{3}yz+3\,{x}^{3}{z}^{2}+3 \,{x}^{2}{y}^{3}+3\,{x}^{2}{y}^{2}z+3\,{x}^{2}y{z}^{2}+3\,{x}^{2}{z}^{ 3}+x{y}^{4}-11\,x{y}^{3}z+3\,...
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1answer
47 views

Find the stronger inequality of $\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ca+2b^{2}+2b}\geq \frac{1}{\sum ab}$

For $a,b,c>0$ and $a+b+c=1.$ Prove$:$ $$\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ca+2b^{2}+2b}\geqq \frac{1}{ab+bc+ca}$$ This inequality is easy and there are two nice proof by AM-GM ...
6
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2answers
88 views

Prove $\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\geqq 1$

For $x,y,z>0$ and $\sqrt{x} +\sqrt{y} +\sqrt{z} =1.$ Prove that$:$ $$\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\geq 1$$ My solution$:$ Let $x=...
4
votes
2answers
100 views

Prove the following inequality $\sum_{i<j<k}\frac{a_ia_ja_k}{(n-2)(n-1)n}\le \bigg(\sum_{i<j}\frac{a_ia_j}{(n-1)n}\bigg)^2+\frac{1}{12}$

For positive integer $n \ge 3$, prove the following inequality $$\sum_{i<j<k}\frac{a_ia_ja_k}{(n-2)(n-1)n}\le \bigg(\sum_{i<j}\frac{a_ia_j}{(n-1)n}\bigg)^2+\frac{1}{12}$$ where $a_1+a_2+\...
3
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3answers
59 views

Proving $(a+b+c)^2\prod_{cyc}(a+b)-4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab\geqq 0$

From Mr. Michael Rozenberg solution: For $a,b,c>0$$,$ prove that$:$ $$(a+b+c)^2\prod_{cyc}(a+b)\geq4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab,$$ I found two SOS proof: 1) $$\text{LHS-RHS}={\frac { \left( ...
1
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2answers
68 views

Prove $\frac{a^2}{(b+c)^2}+\frac{b^2}{(c+a)^2}+\frac{c^2}{(a+b)^2}+\frac{1}{4}\ge \frac{a^2+b^2+c^2}{ab+bc+ca}$

For $a,b,c>0$. Prove that: $$\frac{a^2}{(b+c)^2}+\frac{b^2}{(c+a)^2}+\frac{c^2}{(a+b)^2}+\frac{1}{4}\geqq \frac{a^2+b^2+c^2}{ab+bc+ca}$$ NguyenHuyen gave the following expression$:$ $$\sum \...
3
votes
3answers
73 views

prove that $3(a+b+c) \geq 8(a b c)^{1 / 3}+\left(\frac{a^{3}+b^{3}+c^{3}}{3}\right)^{1 / 3}$

Question - Suppose a,b,c are positive real numbers , prove that $3(a+b+c) \geq 8(a b c)^{1 / 3}+\left(\frac{a^{3}+b^{3}+c^{3}}{3}\right)^{1 / 3}$ (Thailand $2006$) My attempt - we can assume ...
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1answer
73 views

Prove $\frac{3}{2} +\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \leqq \frac{a}{b}+\frac{b}{c} +\frac{c}{a}$

For $a,\,b,\,c>0$. Prove: $$\frac{3}{2} +\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \leqq \frac{a}{b}+\frac{b}{c} +\frac{c}{a}$$ My work: After a lot of caculates, I found: $\text{RHS-LHS}=$ ...
2
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3answers
106 views

Proving $(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}) +\frac{486(ab+bc+ca)^3}{(a+b+c)^6} \geqq 27$

For $a,b,c > 0$ prove: $$(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}) +\frac{486(ab+bc+ca)^3}{(a+b+c)^6} \geqq 27$$ My work: I can easy found SOS for it: $$\text{LHS-RHS}=\sum {\...
3
votes
3answers
80 views

show this inequality $\sum_{cyc}\frac{1}{5-2xy}\le 1$

let $x,y,z\ge 0$ and such $x^2+y^2+z^2=3$ show that $$\sum_{cyc}\dfrac{1}{5-2xy}\le 1$$ try: $$\sum_{cyc}\dfrac{2xy}{5-2xy}\le 2$$ and $$\sum_{cyc}\dfrac{2xy}{5-2xy}\le\sum_{cyc}\dfrac{(x+y)^2}{\frac{...
1
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3answers
71 views

Prove $\Big[\sum\limits_{cyc} a(a^2+2bc)\Big]^3 \geqq 3(ab+bc+ca)^2 . \sum\limits_{cyc} a(a^2+2bc)^2$

For $a,b,c>0$, prove that: $$ \Big[\sum\limits_{cyc} a(a^2+2bc)\Big]^3 \geqq 3(ab+bc+ca)^2 . \sum\limits_{cyc} a(a^2+2bc)^2$$ BW works here, but it's very ugly! My try: Let $p=a+b+c,q=ab+bc+ca,r=...
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4answers
60 views

MOP 2011 inequality

If $a,b,c$ are positive integers prove that $\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +\sqrt{(a^2+c^2-ac)} +9(abc)^{1/3} \le 4(a+b+c)$ My attempt: I tried to split inequality and prove it bit by ...
1
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1answer
71 views

Inequality with 4 variables

For $a,b,c,d>0$ such that $abcd\ge 1$. Prove that $$\frac{1}{3a+2b+c+6}+\frac{1}{3b+2c+d+6}+\frac{1}{3c+2d+a+6}+\frac{1}{3d+2a+b+6} \leq \frac{1}{3}$$ My attempts: By AM-GM $$3a+2b+c+6\geq12\sqrt[...
2
votes
2answers
104 views

Prove that : $m_{a}m_{b}m_{c}\leq\frac{Rs^{2}}{2}$

Let $m_{a},m_{b},m_{c}$ be the lengths of the medians and $a,b,c$ be the lengths of the sides of a given triangle , Prove the inequality : $$m_{a}m_{b}m_{c}\leq\frac{Rs^{2}}{2}$$ Where : $s : \...
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2answers
145 views

Inequality for Olympiad students

Let $a,b,c$ be positive numbers such that $a+b+c=3$. Prove that $\sqrt{3a+\frac{1}{b}}+\sqrt{3b+\frac{1}{c}}+\sqrt{3c+\frac{1}{a}} \ge 6$ My attempts you can see here: https://scontent-xsp1-2.xx....
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4answers
143 views

$\frac{a}{b}+ \frac{b}{c} + \frac{c}{a} \geq \frac{9(a^2+b^2+c^2)}{(a+b+c)^2}$

Prove the following inequality $$\frac{a}{b}+ \frac{b}{c} + \frac{c}{a} \geq \frac{9(a^2+b^2+c^2)}{(a+b+c)^2}, \enspace \forall a,b,c \in (0,\infty)$$ I tried by multying both sides by the ...
1
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3answers
75 views

Given positives $a, b, c$ such that $a + b + c = 3$, prove that $\sum_{cyc}\frac{1}{a^2 + 4b^2 + c^2} \le \frac{1}{2}$.

Given positives $a, b, c$ such that $a + b + c = 3$, prove that $$\frac{1}{c^2 + 4a^2 + b^2} + \frac{1}{a^2 + 4b^2 + c^2} + \frac{1}{b^2 + 4c^2 + a^2} \le \frac{1}{2}$$ We have that $$a^2 + 4b^2 + c^...
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2answers
110 views

For $a,b,c \ge 0$ real numbers, prove $\frac{2}{(1+a)^2}+\frac{2}{(1+b)^2}+\frac{2}{(1+c)^2} \geq \frac{9}{3+ab+bc+ca}$

Let be $a,b,c \ge 0$ real numbers. Prove that: $$\frac{2}{(1+a)^2}+\frac{2}{(1+b)^2}+\frac{2}{(1+c)^2} \geq \frac{9}{3+ab+bc+ca}$$ It is question 2 from here : https://artofproblemsolving.com/...
5
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4answers
164 views

How to prove $\frac a{\sqrt{a^2+3b^2+3c^2}}+\frac b{\sqrt{3a^2+b^2+3c^2}}+\frac{c}{\sqrt{3a^2+3b^2+c^2}}\le\frac3{\sqrt7}$ when $a,b,c>0$

I want to prove that for $a,b,c>0$ we have $$\sum_{cyc} \frac a{\sqrt{a^2+3b^2+3c^2}}= \frac a{\sqrt{a^2+3b^2+3c^2}}+\frac{b}{\sqrt{3a^2+b^2+3c^2}}+\frac{c}{\sqrt{3a^2+3b^2+c^2}}\le\frac3{\sqrt7}.$...
0
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3answers
141 views

A nontrvial inequality $\sum\limits_{\rm cyc}a^2 c(4a-3b-c)^2\ge 20abc\sum\limits_{\rm cyc}a(a-b)$

For real numbers $a,b,c>0$, show that $$\sum_{\rm cyc}a^2 c(4a-3b-c)^2\ge 20abc(a^2+b^2+c^2-ab-bc-ca)$$ This has the following dumbass notation: $$\begin{array}{ccccccccccc} &&&&...
12
votes
3answers
381 views

Hard inequality for positive numbers

The problem is to prove that for $a,b,c>0$ we have $$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{9abc}{4(a^3+b^3+c^3)}\geq \frac{15}{4}.$$ I have tried to use Bergstrom/Engel inequality ...
6
votes
4answers
450 views

If $x,y,z>0.$Prove: $(x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right) \geq9\sqrt[]\frac{x^2+y^2+z^2}{xy+yz+zx}$

If $x,y,z>0.$Prove: $$(x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)\geq9\sqrt[]\frac{x^2+y^2+z^2}{xy+yz+zx}$$ I was not able to solve this problem instead I could solve similar ...
2
votes
1answer
95 views

show this inequality $\left(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\right)^3+12\ge 13(x^3+y^3+z^3)$

let $x,y,z>0$, and such $xyz=1$,show that $$\left(\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\right)^3+12\ge 13(x^3+y^3+z^3)\tag{1}$$ I have konwn use C-S we have $$\left(\dfrac{x^2}{y}+\dfrac{...
1
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1answer
53 views

Finding maximum with two constraints

Let $a$,$b$,$c$ be positive real numbers satisfying $$a+b+c=1$$ $$a^2+b^2+c^2=\frac{3}{8}$$ Find the maximum value of $$a^3+b^3+c^3$$ Using the well known $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-...
0
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0answers
38 views

An elegant but difficult inequality. [duplicate]

$$\frac{x}{\sqrt{x+y}} + \frac{y}{\sqrt{y+z}} + \frac{z}{\sqrt{z+x}}$$ Over all non-negative $x,y,z$ satisfying $x+y+z=4$, let the maximum value of the above expression be $M$. What is the value of $...
1
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1answer
63 views

I can't find the maximum value of P because of some problems in my solution.

Let a,b,c >0 satisfy :a+b+c=3. Find Max P, P= $\frac{2}{3+ab+bc+ca}+\sqrt[3]{\frac{abc}{(1+a)(1+b)(1+c)}}$ I supposed that a=b=c. So that: $\sqrt[3]{\frac{abc}{(1+a)(1+b)(1+c)}}\leq \frac{1}{3}(\...
1
vote
2answers
77 views

show this inequality $\sum_{cyc}\frac{a^3}{a^2+ab+b^2}\ge\sqrt{\sum a^3}$

let $a,b,c>0$.such $a+b+c=1$ show that $$\sum_{cyc}\dfrac{a^3}{a^2+ab+b^2}\ge \sqrt{a^3+b^3+c^3}$$ I have show that not stronger inequality: $$\sum\dfrac{a^3}{a^2+ab+b^2}=\sum_{cyc}\dfrac{a^4}{a^3+...
4
votes
4answers
129 views

Maximizing $\frac{a^2+6b+1}{a^2+a}$, where $a=p+q+r=pqr$ and $ab=pq+qr+rp$ for positive reals $p$, $q$, $r$

Given $a$, $b$, $p$, $q$, $r \in\mathbb{R_{>0}}$ s.t. $$\begin {cases}\phantom{b}a=p+q+r=pqr \\ab =pq+qr+rp\end{cases} $$ Find the maximum of $$\dfrac{a^2+6b+1}{a^2+a}$$ This question is ...
0
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4answers
180 views

Given $x+y+z=3, x,y,z>0 $ how to prove that $\frac{x}{y}+\frac{y}{z}+\frac{z}{x} >= x^2+y^2+z^2$

Given $x+y+z=3, x,y,z>0 $ how to prove that $\frac{x}{y}+\frac{y}{z}+\frac{z}{x} >= x^2+y^2+z^2$ ? I tried something basic like $(x+y+z)^2 = x^2+y^2+z^2 + 2xy+2zx+2yz$, so we just need to ...
12
votes
1answer
455 views

It may be a strengthening form of mean inequality

Let $a_{i}>0,i=1,2,\cdots,n>2$,show that following inequality $$\prod_{i=1}^{n}\left(\dfrac{\displaystyle\sum_{j\neq i}a_{j}}{n-1}\right)^{n-1}\ge\left(\dfrac{\displaystyle\sum_{i=1}^{n}\prod_{j\...
1
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1answer
93 views

$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-3\geq k\left ( \frac{x^{2}+y^{2}+z^{2}}{xy+yz+zx}-1 \right )$

Let $x,y,z>0$. Find the maximum value of $k$ such that the inequality $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-3\geq k\left ( \frac{x^{2}+y^{2}+z^{2}}{xy+yz+zx}-1 \right )$ is true for all $x,y,z>0$....
1
vote
1answer
68 views

Given positive $a, b, c$, prove that $(a^2 + b^2 + c^2)^3 \ge (a^3 + b^3 + c^3)(ab + bc + ca)(a + b + c)$. [duplicate]

Given positive $a, b, c$, prove that $$\large (a^2 + b^2 + c^2)^3 \ge (a^3 + b^3 + c^3)(ab + bc + ca)(a + b + c)$$ As a starting point, $$(a^2 + b^2 + c^2)^3 \ge (a^3 + b^3 + c^3)(ab + bc + ca)(a + ...
5
votes
1answer
115 views

Given three positive numbers $a,b,c$. Prove that $\sum\limits_{cyc}\sqrt{\frac{a+b}{b+1}}\geqq3\sqrt[3]{\frac{4\,abc}{3\,abc+1}}$ .

Ji Chen. Given three positive numbers $a, b, c$. Prove that $$\sum\limits_{cyc}\sqrt{\frac{a+ b}{b+ 1}}\geqq 3\sqrt[3]{\frac{4\,abc}{3\,abc+ 1}}$$ Of course, we've to solve it by $uvw$, before that,...
5
votes
3answers
119 views

Prove $x+y+z \ge xy+yz+zx$

Given $x,y,z \ge 0$ and $x+y+z=4-xyz$ Then Prove that $$x+y+z \ge xy+yz+zx$$ My try: Letting $x=1-a$, $y=1-b$ and $z=1-c$ we get $$(1-a)+(1-b)+(1-c)+(1-a)(1-b)(1-c)=4$$ $$-(a+b+c)-(a+b+c)+ab+bc+...
0
votes
1answer
111 views

Given three positive numbers $x,y,z$ so that $x+y+z=\frac{3}{2}$. Prove that $28\,x^2y^2z^2+3(x^2y^2+y^2z^2+z^2x^2)\leqq1$ .

Problem. Given three positive numbers $x, y, z$ so that $x+ y+ z= \frac{3}{2}$. Prove that $$28\,x^{2}y^{2}z^{2}+ 3(x^{2}y^{2}+ y^{2}z^{2}+ z^{2}x^{2})\leqq 1$$ Remark. It's the cute twin sister of ...
2
votes
1answer
90 views

Solve for $x,y,z \in \mathbb{R^+}$ ,$x^2+y^2+z^2=xyz+4$ and $xy+yz+zx=2(x+y+z)$

Find all positive real numbers such that $$x^2+y^2+z^2=xyz+4$$ And $$xy+yz+zx=2(x+y+z)$$. I substitute $x,y,z$ by $(a+\frac{1}{a}),(b+\frac{1}{b}),(c+\frac{1}{c})$ respectively where $abc=1$ ...
5
votes
1answer
195 views

Given three positive numbers $x,y,z$, prove that $(xyz+x^{2}y+y^{2}z+z^{2}x)^{4}\geqq\frac{256}{27}(x+y+z)^{3}x^{3}y^{3}z^{3}$ .

Hint. Given three length-sides $a, b, c$ of a triangle. For $\Delta= \sqrt{s(s- a)(s- b)(s- c)}$, prove : $$abc+ (a- b)(b- c)(c- a)\geqq \frac{8}{3}\sqrt{\sqrt{3}\Delta^{3}} \tag{1}$$ $$abc- (a- b)...
-1
votes
1answer
111 views

Given numbers $a,b,c\geqq0$ and $-\frac{2}{11}\leqq k\leqq0$. Prove that $(k+1)^{6}(a+b+c)^{2}(\!ab+bc+ca\!)^{2}-81\prod\limits_{sym}(ka+b)\geqq0$ .

Problem. Given three numbers $a, b, c\geqq 0$ and $k= constant$ so that $- \dfrac{2}{11}\leqq k\leqq 0$. Prove that : $$(\!k+ 1\!)^{6}(\!a+ b+ c\!)^{2}(\!ab+ bc+ ca\!)^{2}\!- 81(ka+ b)(kb+ a)(kb+ c)(...
3
votes
1answer
123 views

Calculate the maximum value of $|ab(a^2 - b^2) + bc(b^2 - c^2) + ca(c^2 - a^2)|$ where $a^2 + b^2 + c^2 = 1$.

$a, b, c$ are reals such that $a^2 + b^2 + c^2 = 1$. Calculate the maximum value of $$\large |ab(a^2 - b^2) + bc(b^2 - c^2) + ca(c^2 - a^2)|$$ This problem appeared in a 9th-grade book about ...
0
votes
2answers
145 views

Given three positive numbers $a,b,c$ so that $abc= 1$. Prove $(a-1+\frac{1}{b})(b-1+\frac{1}{c})(c-1+\frac{1}{a})\leqq\frac{2}{a+b+c-1}$ .

(An problem due to Michael Rozenberg) Given three positive numbers $a,\,b,\,c$ so that $abc= 1$. Prove $$\left ( a- 1+ \frac{1}{b} \right )\left ( b- 1+ \frac{1}{c} \right )\left ( c- 1+ \frac{1}{a} \...
2
votes
2answers
158 views

Given three positive numbers $a,\,b,\,c$ . Prove that $(\!abc+ a+ b+ c\!)^{3}\geqq 8\,abc(\!1+ a\!)(\!1+ b\!)(\!1+ c\!)$ .

Given three positive numbers $a,\,b,\,c$ . Prove that $(\!abc+ a+ b+ c\!)^{3}\geqq 8\,abc(\!1+ a\!)(\!1+ b\!)(\!1+ c\!)$ . My own problem is given a solution, and I'm looking forward to seeing a ...
2
votes
1answer
99 views

How do you come up with great and ingenious questions [closed]

In mathematics how are the professionals and great authors able to come up with such ingenious questions which which are so difficult yet elegant. Like in integral calculus it involve a substitution ...
1
vote
3answers
114 views

Minimize value of the function $a^2+b^2+c^2+2\sqrt{3abc}$

Let $a,b,c$ be the positive real numbers such that $a+b+c=1$. Find Minimize of $$P=a^2+b^2+c^2+2\sqrt{3abc}$$ WA says that $P$ gets only a local minimum. But i think it must be maximum value of $P$. ...
2
votes
1answer
131 views

Nice olympiad inequality

Let's go for an olympiad inequality : let $a,b,c>0$ then we have : $$\sum_{cyc}\frac{ab}{a+b}\geq \frac{3\sqrt{3}}{2}\sqrt{\frac{abc}{a+b+c}}$$ My proof : $$\sum_{cyc}\frac{ab}{a+b}=\frac{(a^...
6
votes
1answer
161 views

For $a,b,c,d > 0$ and $abcd = 1$, show that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{12}{a + b + c + d} \geq 7$

Question: For $a,b,c,d > 0$ and $abcd = 1$, show that $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{12}{a + b + c + d} \geq 7$$ My Attempts: Trivial to see that equality ...
3
votes
1answer
211 views

Given three positve numbers $a,b,c$. Prove that $\sum\limits_{cyc}\frac{a}{\sqrt{b(a+b)}}\geqq \sum\limits_{cyc}\frac{a}{\sqrt{b(c+a)}}$ .

Given three positve numbers $a, b, c$. Prove that $$\sum\limits_{cyc}\frac{a}{\sqrt{b(a+ b)}}\geqq \sum\limits_{cyc}\frac{a}{\sqrt{b(c+ a)}}$$ I tried on Holder inequality and https://...
0
votes
3answers
147 views

Find $f(x)$ so that $\sum\limits_{cyc}a^{2}-f(x)\left(\prod_{sym}a-\prod_{sym}(1-a)\right)\geqq3(\frac{x}{2})^{2}$ .

Find $f(x)$ so that $$\sum\limits_{cyc}a^{2}- f(x)\left (\prod\limits_{sym}a- \prod\limits_{sym}(1- a) \right )\geqq 3\left ( \frac{x}{2} \right )^{2}$$ OP. Given three numbers $a, b, c$ so that $\{...
0
votes
1answer
162 views

Given three positive numbers $a,b,c$ so that $8abc\geqq a+b+c+5$. Prove that $\frac{1}{a+2b}+\frac{1}{b+ 2c}+\frac{1}{c+2a}\leqq1$ .

Given three positive numbers $a, b, c$ so that $8abc\geqq a+ b+ c+ 5$. Prove that $$\frac{1}{a+ 2b}+ \frac{1}{b+ 2c}+ \frac{1}{c+ 2a}\leqq 1$$ The best $constant$ here is also $8$, which I found by ...
0
votes
1answer
149 views

Given three positive numbers $a,b,c$. Prove that $\prod\limits_{sym}(\!a+\frac{1}{a}\!)\geqq\frac{4}{3}\sum\limits_{sym}\frac{b+c}{a}$ .

Given three positive numbers $a, b, c$. Prove that $$\prod\limits_{sym}\left(\!a+ \frac{1}{a}\!\right)\geqq \frac{4}{3}\sum\limits_{sym}\frac{b+ c}{a}$$ I use discriminant and uvw and find it with ...
2
votes
1answer
94 views

show this $\sum_{cyc}\frac{x}{x^2-x+1}\le\frac{8}{3}$ [duplicate]

let $x,y,z,w\in R$,and such $x+y+z+w=2$.show that $$\sum_{cyc}\dfrac{x}{x^2-x+1}\le\dfrac{8}{3}$$ I have only solve when $x,y,z,w>0$, because $$\dfrac{x}{x^2-x+1}\le\dfrac{4}{3}x$$ so $$\sum_{...