# Questions tagged [rearrangement-inequality]

Proofs of inequalities by using Rearrangement inequality or Chebyshov inequality.

92 questions
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### For $x+ y+ z= 3$ prove that $4\geqq x^2y+y^2z+z^2x$

Given $3$ positive real numbers $x,\,y,\,z$ and $x+ y+ z= 3$. Prove that $$4\geqq x^2y+ y^2z+ z^2x$$ This problem was homogenized so I set $x+ y+ z= 3$ to cancel stuff. Now I'm stuck. I have noticed ...
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### Prove that $A_1 \cdot B_1 + A_2 \cdot B_2 > A_1 \cdot B_2 + A_2 \cdot B_1$ for $0 < A_1 < A_2$ and $0 < B_1 < B_2$

We are asked to prove or disprove that this is correct $A_1 \cdot B_1 + A_2 \cdot B_2 > A_1 \cdot B_2 + A_2 \cdot B_1$ for $0 < A_1 < A_2$ and $0 < B_1 < B_2$. I'm not very ...
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### Proving that $\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}> \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

Prove that $\dfrac{ab}{c^3}+\dfrac{bc}{a^3}+\dfrac{ca}{b^3}> \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$, where $a,b,c$ are different positive real numbers. First, I tried to simplify the proof ...
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### If $a, b, c$ are sides of a triangle, prove that $\frac{a}{b+c-a} + \frac{b}{a+c-b} + \frac{c}{a+b-c} \ge 3$ [duplicate]

If $a, b, c$ are sides of a triangle, how can I prove that $$\frac{a}{b+c-a} + \frac{b}{a+c-b} + \frac{c}{a+b-c} \ge 3$$
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### Prove the inequality using Chebyshev's Inequality

If $a,b,c \in(0,\infty)$, then prove that: $$9(a^3+b^3+c^3)\ge(a+b+c)^3$$ I was trying to prove this inequality using Chebyshev's Inequality and assuming $a\ge b \ge c$ but to no avail. Can please ...
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### How do I prove this inequality: $a^2+b^2+c^2 \geq ab+bc+ac$? [duplicate]

How do I prove that for any $a, b, c \in \mathbb{R}$ the inequality $a^2+b^2+c^2 \geq ab+bc+ac$ is true?
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Let $n\ge 2$ be postive integer and $a_{i},(i=1,2,\cdots,n)$ be real numbers such that $a_{i}>1,(i=1,2,\cdots,n),\prod_{i=1}^{n}(a_{i}-1)=1$. Show that $$\sum_{i=1}^{n}\dfrac{1}{\displaystyle\... 1answer 100 views ### Prove that \frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y} \geq x^2+y^2+z^2 Let x,y,z \in \mathbb{R} such that x \geq y \geq z > 0. Prove that$$\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y} \geq x^2+y^2+z^2.$$I rearranged the inequality to get$$xy(x^2y)+yz(y^2z)+xz(...
I am learning Inequalities and today I've encountered the famous Rearrangement Inequality. I'll state the theorem: Consider any two collections of real numbers in increasing order, a_1\leq a_2\...