Inequality proof by using the Muirhead inequality.

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### 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 $c_i \neq 0$ ...
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### 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 ...
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### 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 (...
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### 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 : $$f\left(x\right)+f\left(y\right)+f\left(z\right)\ge1$$...
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### $\sum \frac{a}{b+c+d}\le \frac{2\sum a^2}{\sum ab}$ 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 ...
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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 ...
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### 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 ...
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### 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 ...
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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 $... 3 votes 2 answers 167 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) $$P=\frac{a^3}{2a+bc} + \... 1 vote 1 answer 87 views ### Problem in normalization inequalities Let a,b,c>0. Prove that:$$\dfrac{(a+b)^2(b+c)^2(c+a)^2}{abc} \ge \dfrac{64}{27}(a+b+c)^3$$First solution: \bullet Since the inequality is homogeneous, we may normalize a+b+c=3, we need to ... 7 votes 4 answers 151 views ### Prove that: S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)} \leq \frac{1}{3} Let a,b,c>0: Prove that: S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)} \leq \frac{1}{3} My solution: We have: \left[\begin{matrix}\frac{1}{x}+\frac{1}{y} \geq \... 2 votes 1 answer 286 views ### Does Muirhead's Also Work on Cyclic Inequalities? So I am trying to learn how to solve inequalities, and came across the following in one of the problems I was trying to solve: \frac{b^3}{a}+\frac{c^3}{b}+\frac{d^3}{c}+\frac{a^3}{d}\geq ab+bc+... 1 vote 4 answers 157 views ### Let x,y,z are the lengths of sides of a triangle such that x+y+z=2. Find the range of xy+yz+xz-xyz . Let x,y,z are the lengths of sides of a triangle such that x+y+z=2. Find the range of xy+yz+xz-xyz . What I Tried: I have tried by doing (x+y+z)(x+y+z)=2*2 = 4. Also I got 2(xy+yz+xz)+(x^2+... 2 votes 1 answer 97 views ### For any real positive numbers a, b, c, prove that 3(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2) \geq abc(a+b+c)^3 [duplicate] My progress is that I applied Hölder’s for this, 3(a^2b+b^2c+c^2a) \frac{(ab^2+bc^2+ca^2)}{abc} \geq (a+b+c)^3 whereas 3(a^2b+b^2c+c^2a) \frac{(ab^2+bc^2+ca^2)}{abc} = (1+1+1)(a^2b+b^2c+c^2a)(\... 1 vote 3 answers 320 views ### Two inequalities with parameters a,b,c>0 such that ca+ab+bc+abc\leq 4 Let a,b,c>0 be such that bc+ca+ab+abc\leq 4. Prove the following inequalities: (a) 8(a^2+b^2+c^2)\geq 3(b+c)(c+a)(a+b), and (b) \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{a^2b}+\... 2 votes 6 answers 136 views ### How to prove \frac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt{abc}? Give a,b,c>0. Prove that:$$\dfrac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt{abc}.$$My direction: (we have the equation if and only if a=b=c) a^{n+1}+a^nb+a^nc \ge 3a^n\sqrt{abc} ... 3 votes 3 answers 143 views ### Proving \sum_{cyc}\sqrt{a^4+a^2b^2+b^4}\geq \sum_{cyc} a\sqrt{2a^2+bc} for non-negative a, b, c I was trying this question with factorization and other similar methods, Let a, b, c \geq 0. Prove that$$\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] \... 3 votes 2 answers 221 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 ... 2 votes 1 answer 151 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}{... 1 vote 3 answers 100 views ### Inequality 6 deg Fora,b,c\ge 0Prove that $$4(a^2+b^2+c^2)^3\ge 3(a^3+b^3+c^3+3abc)^2$$ My attempt: $$LHS-RHS=12(a-b)^2(b-c)^2(c-a)^2+2(ab+bc+ca)\sum_{sym} a^2(a-b)(a-c)$$ $$+\left(\sum_{sym} a(a-b)(a-c)\right)^2+... 3 votes 0 answers 136 views ### Prove that if abc \geq 1 and a,b,c > 0 then \frac{1}{1 + a + b} + \frac{1}{1 + b + c} + \frac{1}{1 + c + a} \leq 1 [duplicate] Prove that if abc \geq 1 and a,b,c > 0 then,$$\frac{1}{1 + a + b} + \frac{1}{1 + b + c} + \frac{1}{1 + c + a} \leq 1$$Can anyone help me with this problem? Tried AM-GM and Cauchy-Schwarz ... 0 votes 2 answers 68 views ### Prove that \frac{a^2+b^2}{(1-a^2)(1-b^2)} + \frac{b^2+c^2}{(1-b^2)(1-c^2)}+\frac{c^2+a^2}{(1-c^2)(1-a^2)} \geq \frac{9}{2} For a,b,c \in (0,1) such that ab+bc+ca=1 Prove that \frac{a^2+b^2}{(1-a^2)(1-b^2)} + \frac{b^2+c^2}{(1-b^2)(1-c^2)}+\frac{c^2+a^2}{(1-c^2)(1-a^2)} \geq \frac{9}{2} I tried repleace 1 by ab+... 1 vote 2 answers 85 views ### Inequality involving AM-GM but its wierd [duplicate] Let a, b, c be positive real numbers. Prove that \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+ \frac{3\cdot\sqrt{abc}}{a+b+c} \geq 4 Ohk now i know using AM-GM that \frac{a}{b}+\frac{b}{c}+\frac{c}{... -1 votes 3 answers 124 views ### Let a,b,c>0 with \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1. Prove that (a + 1)(b + 1)(c + 1) \geq 64 [duplicate] Let a,b,c>0 with \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1. Prove that (a + 1)(b + 1)(c + 1) \geq 64 Ohk so we are given that abc=a+b+c with that now the inequality becomes 2abc+(a+b+c)+1 \... 3 votes 2 answers 95 views ### Inequality question. Let a,b,c>0 with a+b+c=1. Show that$$\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 &... 2 votes 1 answer 139 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}+\... 6 votes 3 answers 127 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 ... 1 vote 1 answer 68 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 ... 1 vote 1 answer 142 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. ... 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 ... 3 votes 2 answers 438 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 ... 0 votes 1 answer 146 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: ... 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)... 1 vote
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### 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%...
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### 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 ...