# Questions tagged [muirhead-inequality]

Inequality proof by using the Muirhead inequality.

46 questions
1answer
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### Show that $\frac{1}{a^2+b^2+1}+\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}\leq 1$.

Let $a, b, c>0$ s.t. $abc (a+b+c)=3$. Show that $\frac{1}{a^2+b^2+1}+\frac{1}{b^2+c^2+1}+ \frac{1}{c^2+a^2+1}\leq 1$. I have no idea how to start.
1answer
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### Show that $(a^2+3)(b^2+3)(c^2+3)(d^2+3)\geq 256$. [duplicate]

Let $a, b, c, d\geq 0$ s.t. $a+b+c+d=4$. Show that $(a^2+3)(b^2+3)(c^2+3)(d^2+3)\geq 256$. I don't know how can I deconditioned the inequality.
0answers
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### Inequality using lengths of the edges of a triangle

If $a,b,c$ are the lengths of the edges of a triangle, show that: $$\frac {6 (a^2+b^2+c^2)}{a+b+c}\geq \frac {(a+b)^2}{b+c}+\frac {(b+c)^2}{a+c}+\frac {(c+a)^2}{a+b}$$ I have no idea how to start.
2answers
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1answer
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### Inequality from AMM problems section

This is Problem 12024 of AMM. It asks to show that if $x,y,z$ are positive reals, and $xyz=1$, then $(x^{10}+y^{10}+z^{10})^{2}\geq 3(x^{13}+y^{13}+z^{13})$. I could show it for the particular case ...
1answer
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2answers
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### If $ab+bc+ca+abc=4$, then $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq 3\leq a+b+c$

Let $a,b,c$ be positive reals such that $ab+bc+ca+abc=4$. Then prove $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq 3\leq a+b+c$ So high guys im a high schooler trying to solve this inequality. I did a few ...
2answers
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### $(xy+yz+zx)(1+6xyz) \geq 11xyz$

I've came across this inequality: Let $x>$, $y>0$ and $z > 0$ with $x+y+z=1$. Prove that $$(xy+yz+zx)(1+6xyz) \geq 11xyz.$$ I don't know where to take it from, I've tried means ...
1answer
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1answer
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### Wild inequality

I'm trying to solve this elementary inequality, but so far no clue. Can anybody solve it? I tried am/gm inequality on both side, but yield the same result hence no inequality. Also I tried other ...
1answer
192 views

2answers
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### How can I prove the this inequality?

$$a^6+b^6+c^6+3a^2 b^2 c^2 \geq 2(a^3 b^3 + b^3 c^3 +c^3 a^3)$$ $\forall a,b,c \in \mathbb{R}$ Can this be done with just weighted AM-GM?
4answers
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### Prove this inequality: $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$

Let $a$,$b$,$c$ be positive real numbers such that $abc=1$. Prove that $$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$$ I tried various methods. ...
2answers
81 views

5answers
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### if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$

This is supposed to be an application of AM-GM inequality. if $abc=1$, then the following holds true: $a^2+b^2+c^2\ge a+b+c$ First of all, $a^2+b^2+c^2\ge 3$ by a direct application of AM-GM....