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Questions tagged [rearrangement-inequality]

Inequalities related to the rearrangement inequality or Chebyshev inequality

The rearrangement inequality:

For $a_1\le\dots\le a_n$, $b_1\le\dots\le b_n$, and any permutation $\sigma\in S_n$, $$\sum_{i=1}^na_ib_{n-i+1}\le\sum_{i=1}^na_ib_{\sigma(i)}\leq\sum_{i=1}^na_ib_i.$$

Chebyshov inequality:

For any $a_1\le\dots\le a_n$ and $b_1\le\dots\le b_n$, $$n\sum_{i=1}^na_ib_{n-i+1}\le\left(\sum_{i=1}^na_i\right)\left(\sum_{i=1}^nb_i\right)\le n\sum_{i=1}^na_ib_i.$$

194 questions
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If both $a,b>0$, then $a^ab^b \ge a^bb^a$

Prove that $a^a \ b^b \ge a^b \ b^a$, if both $a$ and $b$ are positive.
John anstice
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How prove this inequality $\sum\limits_{cyc}\frac{x+y}{\sqrt{x^2+xy+y^2+yz}}\ge 2+\sqrt{\frac{xy+yz+xz}{x^2+y^2+z^2}}$

let $x,y,z$ are postive numbers,show that $$\dfrac{x+y}{\sqrt{x^2+xy+y^2+yz}}+\dfrac{y+z}{\sqrt{y^2+yz+z^2+zx}}+\dfrac{z+x}{\sqrt{z^2+zx+x^2+xy}}\ge 2+\sqrt{\dfrac{xy+yz+xz}{x^2+y^2+z^2}}$$ My try: Without loss of generality，we assume that…
math110
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Prove that $\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n} \geq (n-1) \left (\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}+\cdots+\frac{1}{\sqrt{x_n}} \right )$

Let $x_1,x_2,\ldots,x_n > 0$ such that $\dfrac{1}{1+x_1}+\cdots+\dfrac{1}{1+x_n}=1$. Prove the following inequality. $$\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n} \geq (n-1) \left…
Jacob Willis
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Proof of the inequality $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$

I am currently attempting to prove the following inequality $\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}$ for all $ a,b,c>0$ My instinctive plan of attack is to use the AM/GM inequality with $x_1=\dfrac{a}{b+c}$ etc. Using that…
Trogdor
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Proving $\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$

The inequality: $$\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$$ Conditions: $a,b,c,d \in \mathbb{R^+}$ I tried using the normal Cauchy-Scharwz, AM-RMS,…
Mathejunior
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Typical Olympiad Inequality? If $\sum_i^na_i=n$ with $a_i>0$, then $\sum_{i=1}^n\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq n$

Let $\sum_i^na_i=n$, $a_i>0$. Then prove that $$ \sum_{i=1}^n\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq n $$ I have tried AM-GM, Cauchy-Schwarz, Rearrangement etc. but nothing seems to work. The fourth power in the LHS really evades me, and I…
Aspirant
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How to prove the equations have only one real solution?

There are $n$ equations. I need answer for the case $n=3$. $$ \frac{1}{x_1}(1-x_1)^2+\frac{1}{x_2}(1-x_2)^2+\cdots+\frac{1}{x_n}(1-x_n)^2=0, $$ and $$ \frac{1}{x_1}(1-x_1^2)^j+\frac{1}{x_2}(1-x_2^2)^j+\cdots+\frac{1}{x_n}(1-x_n^2)^j=0,\; 2\leq…
cbi
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Find a permutation $x_{\sigma(1)},\ldots,x_{\sigma(n)}$ of $x_1,\ldots,x_n$ that maximises $\sum_{k=1}^{n-1}\vert x_{\sigma(k)}-x_{\sigma(k+1)}\vert.$

Suppose $\ x_1,\ x_2,\ \ldots,\ x_n\ $ are real numbers with $\ x_1 < x_2 <\ldots < x_n.$ Is there an efficient way to find a permutation $\ x_{\sigma(1)},\ x_{\sigma(2)},\ \ldots,\ x_{\sigma(n)}\ $ of $\ x_1,\ x_2,\ \ldots,\ x_n\ $ which maximises…
Adam Rubinson
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Geometric Interpretation of Rearrangement Inequality

We know that many of the famous classical inequalities have geometric interpretations. Can you give a geometric interpretation of Rearrangement Inequality? Note: Rearrangement Inequality is $$x_ny_1 + \dots + x_1y_n \leq x_{\sigma(1)}y_1+ \dots +…
scarface
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Proving the inequality $4\ge a^2b+b^2c+c^2a+abc$

So, a,b,c are non-negative real numbers for which holds that $a+b+c=3$. Prove the following inequality: $$4\ge a^2b+b^2c+c^2a+abc$$ For now I have only tried to write the inequality as $$4\left(\frac{a+b+c}3\right)^3\ge a^2b+b^2c+c^2a+abc$$ but I…
arunoruto
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Proving with AM-GM Inequality

$$\frac{4}{abcd}\geq\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}$$ Given: $a+b+c+d=4$ and $a$, $b$, $c$ abd $d$ are positives. How to prove the above inequality using Arithmetic Geometric Mean Inequality? I tried the following but, I am getting…
Mrigank Shekhar Pathak
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If $\sum ab=1$, find the minimum of $\sum\sqrt{\frac a{8ab+1}}$

Let $a$, $b$, $c$ be non-negative real numbers such that $ab+bc+ca=1$. Find the minimum of $$\sqrt{\frac a{8ab+1}}+\sqrt{\frac b{8bc+1}}+\sqrt{\frac c{8ca+1}}.$$ Source: Crux Mathematicorum I tried the cases $a=b=c$ and $a=0$, and I guess that the…
youthdoo
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If $a^2 + b^2 + c^2 = 1$, what is the the minimum value of $\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}$?

Suppose that $a^2 + b^2 + c^2 = 1$ for real positive numbers $a$, $b$, $c$. Find the minimum possible value of $\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}$. So far I've got a minimum of $\sqrt {3}$. Can anyone confirm this? However, I've been…
Boris Poris
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If $a, b, c\in\mathbb R^+, $ then prove that $a^3b+b^3c+c^3a\ge abc(a+b+c) .$

While trying to prove it, I proved the following two inequalities: $a^4+b^4+c^4\ge abc(a+b+c)$ and $(a^2b+b^2c+c^2a)(ab+bc+ca)\ge abc(a+b+c)^2.$ The later one, on some simplification gives $a^3b+b^3c+c^3a\ge abc(ab+bc+ca).$ But we can't claim that…
Dhrubajyoti Bhattacharjee
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Show $ \frac {a^3}{a^2+b^2+c^2-bc}+\frac {b^3}{a^2+b^2+c^2-ca} +\frac {c^3}{a^2+b^2+c^2-ab} \le \frac {(a+b+c)^2 (a^4+b^4+c^4)}{6abc(ab+bc+ca)} $

If $a,b,c$ are positive real numbers then how do we prove that $ \dfrac {a^3}{a^2+b^2+c^2-bc}+\dfrac {b^3}{a^2+b^2+c^2-ca} +\dfrac {c^3}{a^2+b^2+c^2-ab} \le \dfrac {(a+b+c)^2 (a^4+b^4+c^4)}{6abc(ab+bc+ca)} $ Please help. Thanks
user123733
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