Skip to main content

Questions tagged [muirhead-inequality]

Inequality proof by using the Muirhead inequality.

Filter by
Sorted by
Tagged with
1 vote
3 answers
147 views

Prove $\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\ge \sqrt{2(ab+bc+ca)+12}$ when $a+b+c=3.$

Given non-negative real numbers $a,b,c$ satisfying $a+b+c=3.$ Prove that$$\color{black}{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\ge \sqrt{2(ab+bc+ca)+12}.}$$ Equality holds at $a=b=c=1$ or $a=b=0;c=3.$ I ...
Dragon boy's user avatar
1 vote
0 answers
73 views

Proving Muirhead’s inequality by AM-GM

I don't understand the second proof of Muirhead’s inequality in page 10-11 First prove \eqref{constantMuirhead} using AM-GM: Let $(c_i)_{i=1}^n$ be a sequence of real numbers such that $c_i \neq 0$ ...
hbghlyj's user avatar
  • 3,047
2 votes
2 answers
110 views

a,b,c>0 and prove $\frac{a}{a+\sqrt{(a+2b)(a+2c)}}+\frac{b}{b+\sqrt{(b+2a)(b+2c)}}+\frac{c}{c+\sqrt{(c+2b)(c+2a)}}\le \frac{3}{4}.$

Let $a,b,c>0$. Prove that $$\frac{a}{a+\sqrt{(a+2b)(a+2c)}}+\frac{b}{b+\sqrt{(b+2a)(b+2c)}}+\frac{c}{c+\sqrt{(c+2b)(c+2a)}}\le \frac{3}{4}.$$ It is from a book. My tries did not lead to anything ...
TATA box's user avatar
1 vote
2 answers
179 views

Conditions for Muirhead's inequality hold for cyclic sums

I know that Muirhead's inequality apply only for symmetrical sums, but all inequalities with cyclic sums I have seen have the sequence in the greater side majorizing the sequence in the smaller side (...
Tiago Cavalcante's user avatar
-2 votes
2 answers
82 views

Conjecture for the inequality $f\left(x\right)+f\left(y\right)+f\left(z\right)\ge1$ where $f\left(x\right)=\frac{x}{x+\frac{1}{x}+1}$ which seems easy

I come back with an inequality checked with Desmos : Let $x\in R^*$ then define : $$f\left(x\right)=\frac{x}{x+\frac{1}{x}+1}$$ Then do we have : $$f\left(x\right)+f\left(y\right)+f\left(z\right)\ge1$$...
Ranger-of-trente-deux-glands's user avatar
1 vote
2 answers
89 views

An inequality with plenty of summations. [closed]

I found a question, but was unable to solve it. The question is: Let k be a positive integer, and let $x_1, x_2, ..., x_n$ be positive real numbers. Prove that $$\left(\sum_{i=1}^n \frac{1}{1+x_i}\...
TheUnknown1's user avatar
0 votes
2 answers
113 views

To prove the following inequalities of positive rational numbers

I have to prove the following inequalities: $$ a^ab^bc^c \ge \ (\frac{a+b}{2})^{\frac{a+b}{2}} (\frac{c+b}{2})^{\frac{c+b}{2}} (\frac{a+c}{2})^{\frac{a+c}{2}} $$ $$(a+b)^{c}(c+b)^{a}(a+c)^{b} < \...
Sharmi C's user avatar
  • 419
3 votes
1 answer
114 views

$\sum \frac{a}{b+c+d}\le \frac{2\sum a^2}{\sum ab}$ if $\sum a =4$

Let $a, b, c, d$ positive real numbers such that $a+b+c+d=4$. Prove that $$\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}+\frac{d}{a+b+c}\le \frac{2(a^2+b^2+c^2+d^2)}{ab+ac+ad+bc+bd+cd}.$$ My idea is ...
user986772's user avatar
-6 votes
1 answer
98 views

Asking About Other Prove To Use In Solving This Inequality

Let $\ (x,y,z) \ $ be positive real numbers such that $\ \ (x+y)(y+z)(z+x)=8 \ \ $ Prove That $x^3y^3+y^3z^3+z^3x^3+x^2y^2z^2-4xyz>=0$ Muirhead Sols First homogenize, by multiplying by $2$ and ...
Will's user avatar
  • 145
2 votes
2 answers
69 views

Find the largest $t$ such that for all positive $x, y, z$ the following inequality is satisfied

Find the largest $t$ such that for all positive $x, y, z$ the following inequality is satisfied: $(xy+xz+yz) \left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)^2 \geq t$. If there were such an ...
 Alice Malinova's user avatar
2 votes
1 answer
85 views

Prove the inequality $\displaystyle\sum_{s y m} x^{4} y^{2} z \geqslant 2 \sum_{s y m} x^{3} y^{2} z^{2}$

Prove for positive $x,y,z$ the inequality $x^{4} y^{2} z+x^{4} z^{2} y+y^{4} x^{2} z+y^{4} z^{2} x+z^{4} y^{2} x+z^{4} x^{2} y \geqslant 2(x^{3} y^{2} z^{2}+x^{2} y^{3} z^{2}+x^{2} y^{2} z^{3})$. I ...
CatMario's user avatar
  • 398
3 votes
1 answer
172 views

Minimum of $abc$ when $a^3+b^3+c^3\geq a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}.$

Let $a,b,c$ be positive real numbers with $abc=k$ such that the inequality $$a^3+b^3+c^3\geq a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}$$ holds for all $a,b,c$. Find the minimum value of $k$. I found that $...
Oshawott's user avatar
  • 3,966
3 votes
2 answers
255 views

Given the triangle ABC. Let BC = a, AC = b, AB = c. Find the minimum value of the following expression

Given the triangle ABC. Let BC = a, AC = b, AB = c. Find the minimum value of the following expression: a) $$P=\frac{4a}{b+c-a} + \frac{9b}{c+a-b} + \frac{16c}{a+b-c}$$ b) $$P=\frac{a^3}{2a+bc} + \...
Trường Hưng Nguyễn's user avatar
1 vote
1 answer
115 views

Problem in normalization inequalities

Let $a,b,c>0$. Prove that: $$\dfrac{(a+b)^2(b+c)^2(c+a)^2}{abc} \ge \dfrac{64}{27}(a+b+c)^3$$ First solution: $\bullet$ Since the inequality is homogeneous, we may normalize $a+b+c=3$, we need to ...
user avatar
7 votes
4 answers
158 views

Prove that: $S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)} \leq \frac{1}{3}$

Let $a,b,c>0$: Prove that: $S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)} \leq \frac{1}{3}$ My solution: We have: $\left[\begin{matrix}\frac{1}{x}+\frac{1}{y} \geq \...
Leomessi's user avatar
3 votes
1 answer
449 views

Does Muirhead's Also Work on Cyclic Inequalities?

So I am trying to learn how to solve inequalities, and came across the following in one of the problems I was trying to solve: $\frac{b^3}{a}$+$\frac{c^3}{b}$+$\frac{d^3}{c}$+$\frac{a^3}{d}\geq ab+bc+...
Andrew Carratu's user avatar
1 vote
4 answers
209 views

Let $x,y,z$ are the lengths of sides of a triangle such that $x+y+z=2$. Find the range of $xy+yz+xz-xyz$ .

Let $x,y,z$ are the lengths of sides of a triangle such that $x+y+z=2$. Find the range of $xy+yz+xz-xyz$ . What I Tried: I have tried by doing $(x+y+z)(x+y+z)=2*2 = 4.$ Also I got $2(xy+yz+xz)+(x^2+...
Unlimited mathematics's user avatar
2 votes
1 answer
130 views

For any real positive numbers $a, b, c$, prove that $3(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2) \geq abc(a+b+c)^3$ [duplicate]

My progress is that I applied Hölder’s for this, $3(a^2b+b^2c+c^2a) \frac{(ab^2+bc^2+ca^2)}{abc} \geq (a+b+c)^3$ whereas $3(a^2b+b^2c+c^2a) \frac{(ab^2+bc^2+ca^2)}{abc} = (1+1+1)(a^2b+b^2c+c^2a)(\...
Supakorn Srisawat's user avatar
1 vote
3 answers
341 views

Two inequalities with parameters $a,b,c>0$ such that $ca+ab+bc+abc\leq 4$

Let $a,b,c>0$ be such that $bc+ca+ab+abc\leq 4$. Prove the following inequalities: (a) $8(a^2+b^2+c^2)\geq 3(b+c)(c+a)(a+b)$, and (b) $\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{a^2b}+\...
Batominovski's user avatar
  • 49.8k
2 votes
6 answers
139 views

How to prove $\frac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}$?

Give $a,b,c>0$. Prove that: $$\dfrac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}.$$ My direction: (we have the equation if and only if $a=b=c$) $a^{n+1}+a^nb+a^nc \ge 3a^n\sqrt[3]{abc}$ ...
Xiuwei Lee's user avatar
3 votes
3 answers
148 views

Proving $\sum_{cyc}\sqrt{a^4+a^2b^2+b^4}\geq \sum_{cyc} a\sqrt{2a^2+bc}$ for non-negative $a$, $b$, $c$

I was trying this question with factorization and other similar methods, Let $a, b, c \geq 0$. Prove that $$\begin{array}{c} \sqrt{a^4+a^2b^2+b^4}+\sqrt{b^4+b^2c^2+c^4}+\sqrt{c^4+c^2a^2+a^4} \\[4pt] \...
Book Of Flames's user avatar
3 votes
2 answers
235 views

AM/GM inequalities

I need some help to prove this inequality... I guess one can use Jensen's then AM/GM inequalities. Let $x_1, x_2, x_3, x_4$ be non- negative real numbers such that $x_1 x_2 x_3 x_4 =1$. We want to ...
Clifford's user avatar
  • 179
2 votes
1 answer
166 views

Proof of inequality by Muirhead

We have to prove: $$\frac{\sqrt{pq}}{p+q+2r}+\frac{\sqrt{pr}}{p+r+2q}+\frac{\sqrt{pr}}{p+r+2q}\leq\frac{3}{4}$$ By multiplying it all out we get the following equivalent: \begin{align*} 4\sum_{cyc}{...
IMOPUTFIE's user avatar
  • 650
1 vote
3 answers
101 views

Inequality 6 deg

For $a,b,c\ge 0$ Prove that $$4(a^2+b^2+c^2)^3\ge 3(a^3+b^3+c^3+3abc)^2$$ My attempt: $$LHS-RHS=12(a-b)^2(b-c)^2(c-a)^2+2(ab+bc+ca)\sum_{sym} a^2(a-b)(a-c)$$ $$+\left(\sum_{sym} a(a-b)(a-c)\right)^2+...
BestChoice123's user avatar
3 votes
0 answers
145 views

Prove that if $abc \geq 1$ and $a,b,c > 0$ then $\frac{1}{1 + a + b} + \frac{1}{1 + b + c} + \frac{1}{1 + c + a} \leq 1$ [duplicate]

Prove that if $abc \geq 1$ and $a,b,c > 0$ then, $$\frac{1}{1 + a + b} + \frac{1}{1 + b + c} + \frac{1}{1 + c + a} \leq 1$$ Can anyone help me with this problem? Tried AM-GM and Cauchy-Schwarz ...
Maths-Lover's user avatar
0 votes
2 answers
71 views

Prove that $\frac{a^2+b^2}{(1-a^2)(1-b^2)} + \frac{b^2+c^2}{(1-b^2)(1-c^2)}+\frac{c^2+a^2}{(1-c^2)(1-a^2)} \geq \frac{9}{2}$

For $a,b,c \in (0,1)$ such that $ab+bc+ca=1$ Prove that $\frac{a^2+b^2}{(1-a^2)(1-b^2)} + \frac{b^2+c^2}{(1-b^2)(1-c^2)}+\frac{c^2+a^2}{(1-c^2)(1-a^2)} \geq \frac{9}{2}$ I tried repleace $1$ by $ab+...
gsdgr rgsg's user avatar
1 vote
2 answers
89 views

Inequality involving AM-GM but its wierd [duplicate]

Let a, b, c be positive real numbers. Prove that $ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+ \frac{3\cdot\sqrt[3]{abc}}{a+b+c} \geq 4$ Ohk now i know using AM-GM that $ \frac{a}{b}+\frac{b}{c}+\frac{c}{...
Aayam Mathur's user avatar
-1 votes
3 answers
133 views

Let $a,b,c>0$ with$ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1$. Prove that $(a + 1)(b + 1)(c + 1) \geq 64$ [duplicate]

Let $a,b,c>0$ with$ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1$. Prove that $(a + 1)(b + 1)(c + 1) \geq 64$ Ohk so we are given that $abc=a+b+c$ with that now the inequality becomes $2abc+(a+b+c)+1 \...
Aayam Mathur's user avatar
3 votes
2 answers
113 views

Inequality question.

Let $a,b,c>0$ with $a+b+c=1$. Show that $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq 3 + 2\cdot\frac{\left(a^3 + b^3 + c^3\right)}{abc}$$ Ohhhkk. So first off, \begin{align} a^3 + b^3+ c^3 &...
Aayam Mathur's user avatar
3 votes
1 answer
167 views

Schur inequality

Show that for all positive real numbers $a$, $b$ and $c$ such that $abc=1$, the inequality $a+b+c+2a^4+2b^4+2c^4\ge \dfrac{3}{2}\left(a^2\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+b^2\left(\dfrac{1}{a}+\...
user avatar
6 votes
3 answers
133 views

Prove that $ a^2+b^2+c^2 \le a^3 +b^3 +c^3 $

If $ a,b,c $ are three positive real numbers and $ abc=1 $ then prove that $a^2+b^2+c^2 \le a^3 +b^3 +c^3 $ I got $a^2+b^2+c^2\ge 3$ which can be proved $ a^2 +b^2+c^2\ge a+b+c $. From here how can I ...
Chris's user avatar
  • 750
1 vote
1 answer
73 views

Stuck when transforming and solving this

Given abc=1 ( all positive real numbers). Prove that: $$\frac ab + \frac bc + \frac ca +3( \frac ba +\frac cb +\frac ac) \ge 2(a +b +c+\frac 1a+ \frac 1b +\frac1c)$$ My attempt: $$\frac ab + \frac ...
D H's user avatar
  • 63
1 vote
1 answer
144 views

Proving $\sum\limits_{\rm cyc} 1/(a^2 -b+4) \geq3/4$

Suppose $a,b,c\in\mathbb R^+$ with $a+b+c=3.$ Prove that $$\frac{1}{a^2- b+4}+\frac{1}{b^2-c+4}+\frac{1}{c^2- a+4}\geqslant\frac{3}{4}.$$ I tried various approaches, but nothing seems to work. ...
Aditya Ghosh's user avatar
  • 1,162
0 votes
2 answers
75 views

Prove $13\sum(a+b)^5 \geq 16 \sum ab(a+b)(4a^2+4b^2+4ab+c^2)$

Given $a,b,c>0$, prove that $$13[(a+b)^5+(b+c)^5+(c+a)^5] \geq 16[ab(a+b)(4a^2+4b^2+4ab+c^2)+bc(b+c)(4b^2+4c^2+4bc+a^2)+ca(c+a)(4c^2+4a^2+4ca+b^2)]$$ I tried subtracting the RHS from the LFS but ...
Dave Robin's user avatar
3 votes
2 answers
456 views

Barnard and Child inequality exercise

Prove that, $$3(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)≥abc(a+b+c)^3$$ For positive $a,b,c$ The exercises in this book are making me crazy. Any help would be appreciated. My attempts: I opened the LHS ...
Arjun's user avatar
  • 1,152
0 votes
1 answer
151 views

Another asymmetric inequality $5+\frac{3(a^2+2b^2+c^2)}{(a+b)(b+c)} \geq \frac{9(a+b)(b+c)(c+a)}{(a+b+c)(ab+bc+ca)}$

A while ago I conjectured this inequality and its (little) sister on AOPS. Here is another related inequality in the opposite direction which I strongly suspect is true, although I don't have a proof: ...
LHF's user avatar
  • 8,521
2 votes
1 answer
88 views

How to prove $a^3+b^3+c^3+d^3+3\left(a+b+c+d\right) \geq 14+2abcd$ when $a^2+b^2+c^2+d^2=4$

This is a problem from AoPS I can't solve: Let $a,b,c,d\geq0$ with $a^2+b^2+c^2+d^2=4$. How can I prove: $$a^3+b^3+c^3+d^3+3\left(a+b+c+d\right) \geq 14+2abcd$$ My attempt: I try setting $a=2\cos(x)...
user avatar
1 vote
1 answer
69 views

For real numbers $x>0, y>0, z>0$ and $x y z=1 .$ prove that $ x^{6}+y^{6}+z^{6} \geq x^{5}+y^{5}+z^{5} $

For real numbers $x>0, y>0, z>0$ and $x y z=1 .$ Prove that $$ x^{6}+y^{6}+z^{6} \geq x^{5}+y^{5}+z^{5} $$ How to show this? I tried using https://en.wikipedia.org/wiki/Muirhead%...
Blabla's user avatar
  • 857
3 votes
2 answers
219 views

Proof verification for $x^{10}+y^{10}+z^{10}\ge x^9+y^9+z^9$ (where $xyz=1$ and $x,y,z\in \mathbb{R}^+$)

My teacher has shown me the following problem: Problem. Let $x,y,z\in \mathbb{R}_+$ with $xyz=1$. Show that:$$x^{10}+y^{10}+z^{10}\ge x^9+y^9+z^9.$$ I think I solved the problem using Muirhead's ...
Mr. Xcoder's user avatar
  • 1,173
2 votes
2 answers
294 views

Proving $\frac{a}{a^2+2b^2}+\frac{b}{b^2+2c^2}+\frac{c}{c^2+2a^2}\geq 1$ when $a^2+b^2+c^2=a^3+b^3+c^3$

In this answer, @MichaelRozenberg stated the following inequality: Let $a$, $b$ and $c$ be positive numbers such that $a^3+b^3+c^3=a^2+b^2+c^2.$ Then $$\frac{a}{a^2+2b^2}+\frac{b}{b^2+2c^2}+\frac{...
Maximilian Janisch's user avatar
3 votes
4 answers
312 views

If $ab+bc+ca\ge1$, prove that $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$

The following problem is from CHKMO 2018 Problem 1: If $ab+bc+ca\ge1$, prove that $$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$$ I tried to use Cauchy–Schwarz inequality, by ...
Culver Kwan's user avatar
  • 2,785
1 vote
2 answers
65 views

Prove an inequality : $\sum_{cyc}\frac{a^3}{abu+b^2v}\geq \frac{a+b+c}{u+v}$ without Jensen's inequality

I'm interested in the following problem : Let $a,b,c>0$ be the variables and $u,v>0$ be constant then we have : $$\sum_{cyc}\frac{a^3}{abu+b^2v}\geq \frac{a+b+c}{u+v}$$ Rewrrting the ...
Ranger-of-trente-deux-glands's user avatar
0 votes
3 answers
99 views

Prove that $3 a^4+ 2 a^3 b - 3 a^2 b^2 - 2 a b^3 + 3 b^4\geq 0$.

Let $a,b\geq 0$. How can I prove $$3 a^4+ 2 a^3 b - 3 a^2 b^2 - 2 a b^3 + 3 b^4\geq 0$$ ? I try using Schur and Muirhead but they didn't work here because Schur is for three variables and Muirhead ...
user avatar
3 votes
3 answers
73 views

Prove that $\sum_\text{cyc} \frac{ab}{ab+b^2+ca}\le 1$

Hello ladies and gentlemen, here I have another inequality that I am struggling with: Let $a,b,c>0$ Then $$\sum_\text{cyc} \frac{ab}{ab+b^2+ca}\le 1.$$ I try to show $$\frac{ab}{ab+b^2+ca}\le\...
ArtOfProblemSolving's user avatar
4 votes
1 answer
530 views

Inequality in 4 variables Vasc's EV

Let $a,b,c,d\geq0$ satisfying $a+b+c+d=4$ . Prove $$\sqrt{a^3+b^3+c^3+d^3}+2(\sqrt3 -1)abcd\geq\sqrt{3(abc+abd+acd+bcd)}$$ Attempt: $a^3+b^3+c^3+d^3=(a+b+c+d)(a^2+b^2+c^2+d^2-ab-bc-cd-da-ac-bd)+3(...
Sullo's user avatar
  • 613
2 votes
4 answers
139 views

How do we prove this inequality?

Suppose $a,b,c > 0$. Prove that $$\frac{a^2}{b^2} +\frac{b^2}{c^2} + \frac{c^2}{a^2} \geq \frac ab + \frac bc + \frac ca.$$ I've tried multiplying everything by the denominator and then I tried to ...
Jesus's user avatar
  • 1,798
2 votes
3 answers
215 views

Problematic inequality & hint

I would like to ask for hint for proving following inequality: $$x^3(1+x)+y^3(1+y)+z^3(1+z)\geq \frac{3}{4}(1+x)(1+y)(1+z)$$ for all $x>0$, $y>0$, $z>0$ such that $xyz=1$. Generally, I tried ...
Bartek's user avatar
  • 197
0 votes
3 answers
174 views

Generalization of the AM-GM inequality for three variables

Theorem. Let $a,b,c$ be three non-negative real numbers. Then $$a^6+b^6+c^6\geq 3a^2b^2c^2+\frac12 (a-b)^2 (b-c)^2 (c-a)^2.$$ Remark. This Theorem is a generalization of the AM-GM inequality for ...
Maximilian Janisch's user avatar
4 votes
3 answers
274 views

A hard inequality indian olympiad problem [duplicate]

If $x,y,z$ are positive real numbers, prove that: $\left(x+y+z\right)^2\left(yz+xz+xy\right)^2\le 3\left(y^2 + yz + z^2\right)\left(x^2 + xz + z^2\right)\left(x^2 + xy + y^2\right)$. I have been ...
Aditya Saran's user avatar
-1 votes
1 answer
84 views

An inequality with four variables

Prove that $$4[(4;0;0;0)]+12[(2;1;1;0)]\ge 12[(3;1;0;0)]+3[(1;1;1;1)]$$ with $$[(a;b;c;d)]=\frac{1}{4!}\sum_{\sigma\in Sym(4)}x_{\sigma(1)}^{a}\cdot x_{\sigma(2)}^b\cdot x_{\sigma(3)}^c\cdot x_{\...
user600785's user avatar